" /> Rolling Motion Without Slipping

# Rolling Motion Without Slipping

If a wheel rolls w/o slipping, its linear speed and acceleration of the CM relative to the ground are equal to that of a point on the rim of the wheel relative to the CM. The constraint eq (rolling without slipping) is: x, We will solve this problem in two ways: R x Note:. roll without slipping A uniform solid ball is placed at rest on an incline of slope angle θ. The diagrams show the masses (m) and radii (R) of the cylinders. Therefore, the equation of the rotational motion for the disk is: The moment of inertia is a positive scalar; therefore, the angular acceleration vector is parallel to the torque of the friction force. For pure rolling motion (i. Since the disk rolls without slipping. This will enable us to do energy analysis of a variety of rolling objects. Rolling without slipping. Since the disk rolls without slipping. People have observed rolling motion without slipping ever since the invention of the wheel. Rolling Motion + = Rolling without slipping Pure translation Pure rotation about CM Subscribe to view the full document. Which quantity is constant before it rolls without slipping? 1. rolling without slipping; over a straight line; has cusps (points with two tangents) A cycloid is curtate (or contracted) if… it is traced out by… points inside a generating circle (r < R) that is rolling without slipping or; points on the surface of the generating circle that is slipping while rolling with v cm > Rω; does not have cusps. We consider the two cases of rolling without slipping and rolling with slipping. 5 seconds, it stops skidding and start to roll without slipping. a moving wire 3. Here is the first case, the wheel rolls to the right so the rotation is clockwise. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of π/2. An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass:. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even vertical). In which order do the objects reach the bottom of the incline?. Related Links. Derive the governing differential equation of motion. Obviously, the center of the wheel is also the center of the mass, so the linear momentum is p = M ωR. Take moments about the mass center. Rolling motion A special case of rigid body motion is rolling without slipping on a stationary ground surface. Describe the subsequent motion of the mud. Key Takeaways Key Points. If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. Rolling motion w/o slipping is composed of pure translational motion and pure rotational motion. For pure rolling motion (i. For a cylinder of mass $$M$$ and radius $$R$$, rolling motion without slipping down a plane inclined at an angle $$\theta$$ with the horizontal, 1. Its moment of inertia about the central axis is. 25) in the book (Section 12. a bead constraints to move on a fixed vs. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed. The diagrams show the masses (m) and radii (R) of the cylinders. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. The axis of rotation travels with the rolling object. Rolling Motion • The wheel in the figure is rolling to the right without slipping. As the cylinder rolls, point G (center) moves along a straight line. 4-6) Rolling Rotational Inertia Oct. The condition that an object rolls without slipping is that the distance traveled in a line equals the arc length turned out by the wheel. 4 Rolling Without Slipping, Slipping, and Skidding Circular Motion Week 3. Rotational Dynamics: Rolling Spheres/Cylinders is used to calculate the torque that produces rotational motion. of the motion are equal, and the time of the motion is the same for the two components, these two speeds are equal. 5, having been released from rest somewhere along the straight section of the track. People have observed rolling motion without slipping ever since the invention of the wheel. A circular hoop rolls without slipping on a ﬂat horizontal surface. So the condition for rolling without slipping becomes. It also explains the corresponding equations of pure rolling. Together with the results of the experiments of. Here is the first case, the wheel rolls to the right so the rotation is clockwise. Its moment of inertia about the central axis is. This page contains the video Rolling Without Slipping, Slipping, and Skidding. Friction and Rolling Resistance This is static friction if the wheel rolls without slipping. 25 M And A Mass Of 2. c) Every point on the rim of the wheel has a different linear velocity. The kinematic diagram of the spool is shown in Fig. Chapter 9: Rotational Motion Rigid body instead of a particle Rotational motion about a fixed axis Rolling motion (without slipping) Angular Quantities “R” from the Axis (O) Linear and Angular Quantities Kinematical Equations Chapter 10: Rotational Motion (II) Angular Quantities: Vector Rotational Dynamics: t Note: t = F R sinq Note: sign of t Rotational Dynamics: I Rotational Dynamics: I. I want to ask about the direction of frictional force in smooth rolling motion which means the rolling object doesn't slide on the surface. Rolling contains rotational motion and translational motion(straight line motion) Suppose a car is moving and suddenly applies brakes and it starts sort of skidding. If the ball starts at an initial height h above the bottom of the loop, what is the minimum value for h such that the ball just completes the loop-the-loop?. A rod of mass is attached to a spring of stiffness. no slipping) vR cm 2 2 2 2 11 22 cm 2 m I R v h v R m. The wheel will slide forward without turning. Physics 111 Lecture 21 (Walker: 10. Let represent the downward displacement of the center of mass of the cylinder parallel to the surface of the plane, and let represent the angle of rotation of the cylinder about its symmetry axis. What is the minimum value µ 0 of the coefficient of static friction between ball and incline so that the ball will roll down the incline without slipping? Solution by Michael A. naturally it accelerates downward. If the object rolls without slipping, determine if it is a disk, a hoop, or a sphere. Rolling Motion (Without Slipping) When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point. Firstly, we describe the motion of a rigid sphere on a rigid horizontal plane. Rolling Motion Problems. Although rolling wheels are everywhere, when most people are asked “what is the axis of rotation of a wheel that rolls without slipping?”, they will answer “the axle”. (a) Draw the free-body diagram for the ball. Energy is still conserved, but the initial potential energy is now converted into two types of kinetic energy. Verify that Equations (A5) and (A6) are correct dimensionally. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. Which quantity is constant before it rolls without slipping? 1. (b) What is the minimum coefficient of friction required for the sphere to roll without slipping? Solution. 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + = for the touching point at any given time! "Rolling without slipping". Salukvadze optimal solution and ranked Pareto optimal solutions. Physics 111 Lecture 21 (Walker: 10. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of π/2. 0 m/s, determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass; and (c) its total energy. Consider a thin axisymmetric disk with mass and radius that rolls without slipping over a stationary and rough horizontal plane, as illustrated in Figure 1. 3, steady motion is discussed in secs. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed. FRICTIONAL ROLLING PROBLEMS When analyzing the rolling motion of wheels, cylinders, or disks, it may not be known if the body rolls without slipping or if it slides as it rolls. Friction force Minimum coefficient of friction is, Kinetic Energy of Rolling Motion. Professor Lewin derived an equation for the acceleration of an object rolling down a ramp (under pure roll conditions). You may also find it useful in other calculations involving rotation. t and q, w, a vs. General Plane Motion Rolling without slipping. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. Rolling Motion of on fixed incline surface. This motion, though common, is complicated. At which point in the derivation was the assumption of rolling without slipping necessary? Exercise. For many dynamics problems, rolling without slipping means there is a friction force acting on the wheel at the contact point P. What are the conditions for slipping without rolling and rolling without slipping? When a ball or a ring is placed on a surface,it can slip,roll or slip with rollingunder what conditions are each of these executed??It probably has something to do with friction. Consider an object with circular cross section that rolls Rolling as Translation and Rotation Combined along a surface without slipping. Gottlieb Let: µ = coefficient of friction between ball and incline. Motion A bowling ball is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to friction it begins to roll until it rolls without slipping. Angular Acceleration • Definition: - Uniform angular acceleration • ω = ω o + αt - Units - Vector direction • Tangential Acceleration - a t = rα Δ. Why is this good? Because we have relationships between linear and rotational motion quantities, such as v = Rw for the linear and angular speeds, or a = R(alpha) for the linear and angular accelerations. Physics Flash Animations. about its center of mass, rolling without. it won't move F 6 Big yo-yo Since the yo-yo rolls without slipping, the center of mass velocity of the yo-yo must satisfy, v cm = rω. As shown in Fig. The spherical shell started from rest. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. of the general equations of motion of a thin disk rolling without slipping on a horizontal surface, we present results of a simple experiment on the time dependence of the motion that indicate the dominant dissipative power loss to be proportional to the Ω2 up to and including the last observable cycle. rolling friction, which is always present and distinct from either of these other two kinds of frictionfor a rollingobject. When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. A disc of radius R rolls without slipping at speed v. Consider a solid sphere of radius R and mass M rolling without slipping. In the special case of rolling without slipping, the distance. At the bottom of the incline, the speed of the hoop's center-of-mass is v. The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. Of the three forces in the system, two act at that point, so they have no lever arm. 0points A bicycle wheel of radius 0. UY1: More About Rolling Motion. The condition for rolling without slipping is The torque is. Another key is that for rolling without slipping, the linear velocity of the center of mass is equal to the angular velocity times the radius. 10-9 Rotational Plus Translational Motion; Rolling In (a), a wheel is rolling without slipping. Rolling, Torque, and Angular Momentum Rolling Motion: • A motion that is a combination of rotational and translational motion, e. No energy is dissipated and Mechanical Energy is Conserved. • Describing rotational motion • Rolling without slipping • Torque Lecture 24: Rotational Motion. The qualitative features of the motion can now be deduced from the equations. People have observed rolling motion without slipping ever since the invention of the wheel. see ISM 10. Kinetic Energy The total KE of an object undergoing rolling motion is the The total KE of an object undergoing rolling motion is the sum of the rotational KE about its CM and the translational sum of the rotational KE about. rolling without slipping. When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. การเคลื่อนที่แบบการกลิ้ง (Rolling Motion, without slipping) หรือ การหมุนรอบแกนที่เคลื่อนที่. For pure rolling motion (i. A wheel rolls uniformly on level ground without slipping. The Kinematics of a Translating and Rotating Wheel model displays the model of wheel rolling on a floor. The motion of any point on the ring can be considered as vector sum of translational and rotational motion. An object is rolling, so its motion involves both rotation and translation. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. This motion, though common, is complicated. In contrast, it is well known that, for rolling without slipping, a uniform cylinder with moment of inertiaI = kma2 about its axis has acceleration gsinα/(1 +k) down an. If the object rolls without slipping, determine if it is a disk, a hoop, or a sphere. The rolling-without-slipping kinematic condition for figure 5. 84, there are three forces acting on the cylinder. Rolling Without Slipping. In order to start the rolling motion, a force or torque must be applied to the wheel. The spool rolls to the right. Which quantity is constant before it rolls without slipping? 1. 4 and 5, and oscillation about steady motion is considered in sec. This will enable us to do energy analysis of a variety of rolling objects. A cylinder of radius rolls without slipping down a plane inclined at an angle to the horizontal. Since the disk rolls without slipping. t for this ride. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (October 2, 2014; updated January 12, 2018) 1Problem Discuss the motion of a cylinder that rolls without slipping on another cylinder, when the latter rolls without slipping on a horizontal plane. Rolling motion is a combination of translational motion and rotational motion. The direction of its angular momentum is _____________. David Pritchard, Prof. see ISM 10. You can view a realistic animation of the rolling with slipping and watch as it changes to pure rolling without slipping. Now the translational kinetic energy of the ring is,Rotational kinetic energy of the ring is,. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. Salukvadze optimal solution and ranked Pareto optimal solutions. Friction force Minimum coefficient of friction is, Kinetic Energy of Rolling Motion. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. see ISM 10. or spherical shell) having mass M, radius R and rotational inertia I. The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. Consider a. Which one of the following is necessarily true? a) All points on the rim of the hoop have the same speed. rotational motion. A circular hoop rolls without slipping on a ﬂat horizontal surface. When discussing the physics of a rolling wheel in an introductory course, it is always fun to point out that a point on the rim of the wheel traces out a cycloid, as shown in Figure 1. to the right 2. Rolling Motion (Without Slipping) When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point. This has 1 less dof Pick the coordinates as shown. Rolling Motion of on fixed incline surface. Determine the natural frequency The statement that the "cord does not slip on the pulley" introduces the rolling-without-slipping condition. 4 – A vertical force but a horizontal motion A spool of mass M has a string wrapped around its. They both travel a sh01t distance at the bottom and then start up another incline. Practice comparing the rotational kinetic energy of two objects based on their shape and motion. For example, an object, say a ball is in rolling, that is it is rolling on the surface of the ground. Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point. Consider the following example. 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + = for the touching point at any given time! "Rolling without slipping". For analyzing rolling motion in this chapter, refer to Figure 10. One disk rolls without slipping, and one slides down the ramp. The diagrams show the masses (m) and radii (R) of the cylinders. 0points A bicycle wheel of radius 0. Chapter 2 Rolling Motion; Angular Momentum 2. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. UY1: More About Rolling Motion. Answer and Explanation: Rolling without slipping is a combination. Consider the following three objects, each of the same mass and radius: a solid sphere a solid disk a hoop All three are released from rest at the top of an inclined plane. At this moment the ratio between magnitudes of velocities of A and B with respect to ground is. A piece of mud on the wheel flies off when it is at the 9 o’clock position (rear of wheel). In this work we analyse the motion of a sphere that rolls without slipping on the inside of a right circular cone under the inﬂuence of a uniform gravitational ﬁeld acting verically downwards, in the direction of the symmetry axis of the cone. (a) Find the linear acceleration of the center of mass. , a ball, cylinder, or disk rolling without slipping. (a) Draw the free-body diagram for the ball. rotational motion. To define such a motion we have to relate the translation of the object to its rotation. From what minimum height above the bottom of the track must the marble be released in order not to leave the track at the top of the loop. David Litster, Prof. Q8: Sketch crude graphs of x, v, a vs. At a given point in time, we can apply the rotational version of Newton’s second law to rotations about the point where the cylinder touches the surface (as the cylinder is rolling without slipping, this is the only motion at that point). Label the identification points above. torque about the center from the force of tension. "Rolling resistance " which slows the wheel is a completely different force. At the bottom of the incline, the speed of the hoop's center-of-mass is v. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even vertical). The motion of a disk that is spun on a smooth flat surface slowly damps out due to friction. Energy is still conserved, but the initial potential energy is now converted into two types of kinetic energy. Consider two cylinders that start down identical inclines from rest except that one is frictionless. For analyzing rolling motion in this chapter, refer to Figure 10. In this section, we compare two situations: (1) rolling and slipping, and (2) rolling without slipping. 4 – A vertical force but a horizontal motion A spool of mass M has a string wrapped around its. these aspects should be properly darified. If the pulley makes four revolutions without the rope slipping, what length of. 26A is Without slipping, the disk has two variables, X and β, but only one degree of freedom. Physics Flash Animations. In other words, at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. Salukvadze optimal solution and ranked Pareto optimal solutions. To define such a motion we have to relate the translation of the object to its rotation. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. It is known that this motion can be considered as a rotation about an. rolls without slipping on a horizontal terrain in such a way that it touches the terrain on the surface line parallel to the y-axis. • The red curve shows the path (called a cycloid) swept out by a point on the rim of the wheel ω swept out by a point on the rim of the wheel. Rolling Motion + = Rolling without slipping Pure translation Pure rotation about CM Subscribe to view the full document. 4 m/s rolls up a hill without slipping. • Describing rotational motion • Rolling without slipping • Torque Lecture 24: Rotational Motion. Iclicker #1. Consider a thin axisymmetric disk with mass and radius that rolls without slipping over a stationary and rough horizontal plane, as illustrated in Figure 1. This friction force prevents slipping. 80 kg climbs an incline. When the ring rolls without slipping, then v = rω, where v is the velocity of centre of mass of ring, and ω is its angular velocity. The condition for rolling without slipping is The torque is. Wednesday, January 23, 2008. In this work we analyse the motion of a sphere that rolls without slipping on the inside of a right circular cone under the inﬂuence of a uniform gravitational ﬁeld acting verically downwards, in the direction of the symmetry axis of the cone. The diagrams show the masses (m) and radii (R) of the cylinders. A solid cylinder W of mass mand radius r, r Rω; does not have cusps. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. 5, having been released from rest somewhere along the straight section of the track. "Rolling resistance " which slows the wheel is a completely different force. There are several assumptions to simplify the analysis of rolling objects: Homogeneous rigid body; High degree of symmetry (E. from rest and rolls without slipping a distance of 6 m If object is rolling with a com≠ 0 (i. For example, we can look at the interaction of a car’s tires and the surface of the road. Hey guys! In this video we're going to talk about conservation of energy in rolling motion problems. A mathematical model of the dynamics of a rolling rigid sphere in a rectilinear chute-conveyor with transformed dry friction is developed. The spherical shell started from rest. The wheel will slide forward without turning. The sphere has two kings of kinetic energy: linear and rotational. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. 22 m and a mass of 1. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. 0points A bicycle wheel of radius 0. 1 Rolling Without Slipping When a round, symmetric rigid body (like a uniform cylinder or sphere) of radius R rolls without slipping on a horizontal surface, the distance though which its center travels (when. The linear speed of the wheel is. Now, consider a wheel of radius R rolling without slipping along the straight. In other words, at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. A circular hoop rolls without slipping on a ﬂat horizontal surface. Derive the governing differential equation of motion. (b) What is the minimum coefficient of friction required for the sphere to roll without slipping? Solution. In which order do the objects reach the bottom of the incline?. Rolling without slipping. A multi-objective optimization problem of a passively controlled motion is formulated for the model and it has been solved. The equations of motion will be: F x = m(a G) x => P - F = m a G F y = m(a G) y => N - mg = 0 M G = I Ga => F r = I G a There are 4 unknowns (F, N, a, and a G. Solution A. When the ring rolls without slipping, then v = rω, where v is the velocity of centre of mass of ring, and ω is its angular velocity. there are net forces) NON-smooth rolling motion. Course Material Related to This Topic: Complete exam problem 1e; Check solution to exam Problem 1e. Multiple-Choice Homework 21: Torque and rolling motion Problem 1: A solid cylinder and a solid sphere are rolling without slipping down an inclined plane. General Plane Motion Rolling without slipping. Velocities of points on a rolling wheel. Theoretical background Consider a cylinder of massm and radius R rolling down an inclined plane of angle θ. 7 in Section 10. Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. could be a cylinder, hoop, sphere. along a surface without slipping. a wheel rolling down the road. The motion of an extended rigid body can be resolved into two parts. Rolling Motion without Slipping. If the conditions are such that the object rolls without slipping, there is also a geometric relationship between the displacement of the center of mass and the angle turned or, equivalently, between the accelerations: x =!. Rolling without slipping commonly occurs when an object such as a wheel, cylinder, or ball rolls on a surface without any skidding. Another key is that for rolling without slipping, the linear velocity of the center of mass is equal to the angular velocity times the radius. A Rolling Object Accelerating Down an Incline. Let m be the mass and r be the radius of ring. b) All points on the rim of the hoop have the same linear velocity. Absolute Motion: Exercise A wheel of radius r rolls without slipping. Rolling without slipping. The kinematic diagram of the spool is shown in Fig. Take moments about the mass center. We otation Combin can simplif ed y its study by treating it as a combination of translation of the center of mass and rotation of the object about the center of mass Consider the two snapshots of a rolling bicycle wheel shown in the figure. 26A is Without slipping, the disk has two variables, X and β, but only one degree of freedom. The angular momentum is non-zero and constant for one of the restaurant's customers seated near a window. This has 1 less dof Pick the coordinates as shown. Iclicker #1. Key ideas for rolling: Rolling can be considered to be a superposition of a pure translational motion and a pure rotational motion. In contrast, it is well known that, for rolling without slipping, a uniform cylinder with moment of inertiaI = kma2 about its axis has acceleration gsinα/(1 +k) down an. Consider an object with circular cross section that rolls Rolling as Translation and Rotation Combined along a surface without slipping. Dorbolo, S. This is defined by motion where the point of contact with the ground has zero velocity, so it matches the ground velocity and is not slipping. What is the minimum value µ 0 of the coefficient of static friction between ball and incline so that the ball will roll down the incline without slipping? Solution by Michael A. The spring is attached to a wall and the rod rests on a disk of the same mass. This motion, though common, is complicated. Rolling involves both of these at the same time - rotation while the wheel is experiencing straight-line motion. Starting Rolling Motion. If the pulley makes four revolutions without the rope slipping, what length of. In fact, the wheel is experiencing both a translational and rotational motion. A mathematical model of the dynamics of a rolling rigid sphere in a rectilinear chute-conveyor with transformed dry friction is developed. • Will only consider rolling with out slipping. v bottom =0 R v CM Quick Quiz and Demo ! Consider a wheel in pure. We consider the two cases of rolling without slipping and rolling with slipping. People have observed rolling motion without slipping ever since the invention of the wheel. the rolling motion of a cylinder or hoop. rotational motion. We can simplify its study by treating it as a combination of translattreating it as a combination of translation of the center ofion of the center of. In rotations, rolling without slipping is the phrase we always hope we see in a problem. David Litster, Prof. A ball slipping on a surface is decelerated by the friction force opposing the ball linear displacement. 11-4 Combining Rolling and Newton’s Second Law for Rotation Let’s now look at how we can combine torque ideas with rolling-without-slipping concepts. The radius of gyration of the disk about point o is kog; hence,. we consider rolling with and without slipping and establish the conditions for the frictional force to have the direction of the centre-of-mass velocity, contrary to the common idea that this force is always opposite to motion. This situation can be analyzed using relative velocity and acceleration equations. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. In our analysis of the motion of a rigid disk on a horizontal plane we adopt a vectorial approach as advocated by Milne [10]. 7 in Section 10. The linear kinetic energy is given by the usual equation: 1/2 m*v^2. The acceleration of the body is , where k is the radius of gyration of body. Determine the natural frequency The statement that the "cord does not slip on the pulley" introduces the rolling-without-slipping condition. and Since the maximum static friction force , it imply that for the disk to rolling without slipping. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. could be a cylinder, hoop, sphere. When objects roll along a surface, like this tire rolls across the ground, there must be some sort of force to cause the rotation to occur. This accelleration is constant until the ball rolls without slipping, which is to say that the angular velocity times the radius matches the velocity of the ball. Rolling without slipping commonly occurs when an object such as a wheel, cylinder, or ball rolls on a surface without any skidding. Rolling Without Slipping. In this work we investigate the motion of a homogeneous ball rolling without slipping on uniformly rotating horizontal and inclined planes under the action of a constant external force. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of π/2. Motion A bowling ball is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to friction it begins to roll until it rolls without slipping. rolling without slipping; over a straight line; has cusps (points with two tangents) A cycloid is curtate (or contracted) if… it is traced out by… points inside a generating circle (r < R) that is rolling without slipping or; points on the surface of the generating circle that is slipping while rolling with v cm > Rω; does not have cusps. A ball with an initial velocity of 8. In short, this study suggests that the new position sensor can accurately sense the position of a cylinder that rolls without slipping. b) All points on the rim of the hoop have the same linear velocity. The kinematic diagram of the spool is shown in Fig. In our analysis of the motion of a rigid disk on a horizontal plane we adopt a vectorial approach as advocated by Milne [10]. Pure Roll of Hollow and Solid Cylinders: In pure roll, the object is not skidding or slipping, and the speed of the center of mass equals the circumferential speed. Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point. So, in this case, the object tries to slide first, which friction prevents, and that leads to rolling. no slipping) vR cm 2 2 2 2 11 22 cm 2 m I R v h v R m. Users can change the type of object (solid sphere, solid cylinder, etc. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity. A and B are two points on the ring. Rolling Without Slipping. For this consider a body with circular symmetry for example cylinder, wheel, disc , sphere etc. physics 111N 2 rotations of a rigid body! suppose we have a body which rotates about some axis rolling "rolling without slipping". When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move.


# Prime Time

We spotted this photograph of a letter to The Telegraph, shared by Card Colm on Twitter earlier in the year. It’s exactly the kind of mathematical claim we like to enjoy verifying, so we thought we’d dig in.

SIR – As one obsessed with prime numbers, I note that we have gone from 2017 (a prime) into 2018 (two times a prime), which will be followed by 2019 (three times a prime). I believe this sequence has only happened three other times in the past 1,129 years.
Keith Burgess-Clements
Maidstone, Kent

It’s lovely that newspapers will print this kind of letter, and a quick check verifies that Mr Burgess-Clements is indeed correct that these three numbers have the properties described:

• $2017$ prime, $2018 = 2 \times 1009$, $2019 = 3 \times 673$

His follow-up statement, that this sequence has only happened three other times in the last 1,129 years, takes a little more checking. But only a little – as we have a resident CL-P, who describes the necessary calculation as ‘a trivial amount of Python code’, and quickly came up with the following list:

• $13$ prime, $14 = 2 \times 7$, $15 = 3 \times 5$
• $37$ prime, $38 = 2 \times 19$, $39 = 3 \times 13$
• $157$ prime, $158 = 2 \times 79$, $159 = 3 \times 53$
• $541$ prime, $542 = 2 \times 271$, $543 = 3 \times 181$
• $877$ prime, $878 = 2 \times 439$, $879 = 3 \times 293$
• $1201$ prime, $1202 = 2 \times 601$, $1203 = 3 \times 401$
• $1381$ prime, $1382 = 2 \times 691$, $1383 = 3 \times 461$
• $1621$ prime, $1622 = 2 \times 811$, $1623 = 3 \times 541$

Here’s that Python code, in case you’re curious. It uses Sage’s Primes() function.

pr = set([p for p in range(2018) if p in Primes()])double = set(2*p-1 for p in pr) triple = set(3*p-2 for p in pr)years = sorted(pr & double & triple)

This is a full list of all the cases below 2017, and hints at some nice more interesting patterns – $13$ and $541$ both occur as the prime year and the prime factor that's a third of another year. But we don't have time to dig into that now – we have to check if a person in the newspaper was wrong!

Keith's claim that this has happened thrice in the last 1,129 years is indeed correct – $2018 – 1129 = 889$, and three sets of years have occurred since then. I suspect this may have been a typo though, as if he'd said "the last 1,139 years", that would have included the tail end of the set starting in $877$. Maybe he was looking for the most impressive length of time with the fewest occurrences, to illustrate how rare it is (in which case 1139 would be your best bet, and probably what he meant). We favour "only four times since 1000AD" which still sounds pretty good.

One final question to answer is, how many of these will there be going forward? The next few will be:

• $2557$ prime, $2558 = 2 \times 1279$, $2559 = 3 \times 853$
• $2857$ prime, $2858 = 2 \times 1429$, $2859 = 3 \times 953$
• $3061$ prime, $3062 = 2 \times 1531$, $3063 = 3 \times 1021$

It's also worth checking when this pattern will continue for four years, so that the fourth year is four times a prime; that's $12721$, which is prime, while $12722 = 2 \times 6361$, $12723 = 3 \times 4241$ and $12724 = 4 \times 3181$.

• #### Katie Steckles

Publicly engaging mathematician, Manchester MathsJam organiser, hairdo.

### 5 Responses to “Prime Time”

1. John

If we write dates as 8 digit numbers yyyymmdd, then 2018 has 18 prime dates. 2019 will have 19 prime dates, and 2021 will have 21.

2. Rafa

Just a Haskell version of the code:

import Data.List
import Math.NumberTheory.Primes.Sieve -- arithmoi library
import qualified Data.Set as Set

years =  Set.intersection (Set.fromList pr) (Set.intersection doubles triples)
where
pr = takeWhile (< 2018) primes
doubles = Set.fromList [2*p-1|p<-pr]
triples = Set.fromList [3*p-2|p<-pr]

main = print years


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# In how many ways can 5 different colored marbles be placed in 3 distin

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In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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14 Oct 2013, 08:53
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In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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15 Oct 2013, 02:19
11
2
How can we fill three pockets with 5 marbles?
3 + 1 + 1
2 + 2 + 1

With 3,1,1 distribution:
# of ways to select 3 from 5
5!/3!2! = 10
# of ways to select 1 ball from 2
2!/1! = 2
# of ways to select 1 ball from 1
1!/1! = 1
How many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 3

10*2*3 = 60

With 2,2,1 distribution:
How many ways to select 2 from 5?
5!/2!3! = 10
# of ways to select 2 from 3
3!/2!1! = 3
# of ways to select 1 from 1
1
How many ways to distribute 2,2,1 to 3 boxes? 3!/2! = 3

10*3*3=90

90+60 = 150

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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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20 Sep 2014, 13:22
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2
This is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )

Anyway this is how I solved it...
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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08 Aug 2015, 07:18
1
1
praffulpatel wrote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

(A) 60
(B) 90
(C) 120
(D) 150
(E) 180

We have 5 marbles and 3 pockets
So we have two cases

Case-1: One Pocket with 3 marbles and two pockets with 1 marble each
No. of Arrangements = 5C3 * 3C1 * 2! = 10*3*2 = 60
5C3 - No. of ways of choosing 3 out of 5 marbles which have to go in one pocket
3C1 - No. of ways of choosing 1 out of 3 pockets in which 3 marbles have to go
2! - No. of ways of arranging remaining 2 marbles between remaining two pockets which get one marble each

Case-2: Two Pockets with 2 marbles each and one pockets with 1 marble
No. of Arrangements = 5C2 * 3C2 * 3C2 = 10*3*3 = 90
5C2 - No. of ways of choosing 2 out of 5 marbles which have to go in one pocket
3C2 - No. of ways of choosing 2 out of remaining 3 marbles which have to go in second pocket
3C2 - No. of ways of choosing 2 out of 3 pockets in each of which 2 marbles have to go

Total cases = 60+90 = 150

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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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09 Aug 2015, 14:23
GMATinsight, Bunuel, VeritasPrepKarishma,

I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.

Then we come to the 3-1-1 and 2-2-1 combinations.

For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.

Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.

Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.

Could you please explain why we ignored 1-1-1 combination?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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10 Aug 2015, 00:18
1
jhabib wrote:
GMATinsight, Bunuel, VeritasPrepKarishma,

I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.

Then we come to the 3-1-1 and 2-2-1 combinations.

For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.

Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.

Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.

Could you please explain why we ignored 1-1-1 combination?

Hey jhabib,

You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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10 Aug 2015, 04:21
1
Hi jhabib

The Question has clearly specified that "ALL the Marbles have to be assigned to 3 pockets" so 1-1-1 needs to be ignored

Along with 1-1-1, 1-1-2, 1-2-1 ans 2-1-1 also need to be ignored.

I hope it helps!

jhabib wrote:
GMATinsight,

I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.

Then we come to the 3-1-1 and 2-2-1 combinations.

For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.

Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.

Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.

Could you please explain why we ignored 1-1-1 combination?

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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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06 Jan 2018, 13:02
2
praffulpatel wrote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

(A) 60
(B) 90
(C) 120
(D) 150
(E) 180

hi

I have seen a solution to a problem similar to this one elsewhere on the forum
let me explain it to you

5 different colored marbles can be placed in 3 distinct pockets without any restriction is

= 3^5 = 243

as we are asked to find out the ways in which any pocket must get at least 1 marble, lets find out the ways in which any pocket must not get at least 1 marble, and then subtract the number of ways in which any pocket must not get at least 1 marble from the total number of ways, that is 3^5

so lets get going

number of ways in which all marbles can get to 1 pocket is

= 3, as there are 3 distinct pocket in total

now, number of ways in which 2 pockets can get all the marbles and 1 pocket remains empty is

= (2^5 - 2) * 3 = 90

2 has been subtracted to eliminate the possibility that any pocket out of 2 can get all the marbles

3 has been multiplied with the whole expression, because 2 pockets out of 3 have been selected

now we are in business

= 243 - 90 - 3

= 150 (D)

hope this helps
thanks

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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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09 Oct 2018, 23:13
jhabib wrote:
GMATinsight, Bunuel, VeritasPrepKarishma,

I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.

Then we come to the 3-1-1 and 2-2-1 combinations.

For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.

Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.

Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.

Could you please explain why we ignored 1-1-1 combination?

Hey jhabib,

You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.

Dear Karishma ..
I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .
All the approaches are considering 3 bags as identical , but in problem they are distinct .
My approach to this problem :
I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .
So first distribute 3 balls one ball each to each bag .
select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .
so total 60 ways .
Now we are left with 2 balls and 3 bags each containing one ball each .
for first ball we have 3 options and second ball again 3 options . 3*3 ways .
TOTAL = 60*3*3=540 ways .

Please somebody help me in this . This problem just made me mad ..
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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10 Oct 2018, 00:22
karnaidu wrote:
Dear Karishma ..
I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .
All the approaches are considering 3 bags as identical , but in problem they are distinct .
My approach to this problem :
I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .
So first distribute 3 balls one ball each to each bag .
select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .
so total 60 ways .
Now we are left with 2 balls and 3 bags each containing one ball each .
for first ball we have 3 options and second ball again 3 options . 3*3 ways .
TOTAL = 60*3*3=540 ways .

Please somebody help me in this . This problem just made me mad ..

You have some double counting here. When you select some from a group and distribute and then distribute the rest to the same bags/pockets, there is double counting.

Say you distributed a, b, c first such that
Then you distributed d and e such that Bag1 got d and bag2 got e.

Take another case.
Say you distributed d, b and c first such that
Then you distributed a and e such that Bag1 got a and bag2 got e.

Note that both cases have exactly the same end result. But you would count them as two separate cases. Similarly, there will be other cases which will be double counted. Hence this method will not work.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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10 Oct 2018, 00:53
sxyz wrote:
This is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )

Anyway this is how I solved it...

Hii ..
In this approach you are missing arrangement of bags . You have to take care of that too ..
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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18 Oct 2018, 11:16
Hi. I don't understand why this has been multiplied by 3!/2! for 2-2-1 combination. Plus, shouldn't 5c2*3c2 be multiplied by 2! as the pockets are distinct, and the arrangement would matter? Thanks
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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05 May 2019, 05:41
sxyz wrote:
This is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )

Anyway this is how I solved it...

Hi, I have a doubt.

Here, the case {3,1,1} = $$5C3*2C1*1C1 = 10*2*1 = 20$$
And this is multiplied by 3 (coz three different pockets.)

But generally, in case of distribution between identical pockets/bags/groups etc., the value is divided by the similar ones.
Such as, in here, {3,1,1} has 2 similarly filled pockets, and hence, divided by 2!.

Is it not being done here like that because the pockets are all different here?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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05 May 2019, 13:48
Hi,

Can we solve it the following way?

Total number of possible combinations: $$3^{5}=243$$
Combinations that do not satisfy given condition (at least 1 marble in each pocket): $$4-1-0$$, $$3-2-0$$ and $$5-0-0$$.
1) 4-1-0. 4 marbles can be allocated to 3 different pockets, the rest - 1 - can also be allocated to 3 different pockets. Thus, number of unsatisfactory combinations in for this case = $$3*3*5 = 45$$ - multiply by 5, since marbles can also differ between the pockets.
2) 3-2-0. The same holds true here = $$3*3*5 = 45$$.
3) 5-0-0. This scenario is easy = $$3$$ different cases are possible.

Thus, $$243-45-45-3=150$$.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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26 Aug 2019, 13:19
praffulpatel wrote:
In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

(A) 60
(B) 90
(C) 120
(D) 150
(E) 180

given: 5 dif marbs 3 dif pockets with at least 1 marble each

$$[3,1,1]…5c3•2c1•1=10•2=20…•arrangements[3,1,1]=20•(3!/2!)=60$$ (3!=num.pockets; 2!=num.duplicates [1,1])
$$[2,2,1]…5c2•3c2•1=10•3=30…•arrangements[2,2,1]=30•(3!/2!)=90$$ (3!=num.pockets; 2!=num.duplicates [2,2])
$$total.arrangements=60+90=150$$

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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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03 Oct 2019, 10:28
We have two cases: The first case has one pocket with 3 marbles, 1 marble in each of the two remaining pockets. The second case has two pockets with 2 marbles each, and one pocket with one marble. We always add cases together to obtain the total number of possibilities.

Case 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1

Case 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.

Thus there are 60+90 = 150 ways we can place the marbles
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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04 Oct 2019, 22:44
praffulpatel wrote:
In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?

(A) 60
(B) 90
(C) 120
(D) 150
(E) 180

Solved it a little differently than others from what I read . So will share.

I first made sure that each of the pocket has at least 1 marble.
That is 5C3 * 3!

Then you can either distribute the remaining 2 as 2, 0, 0 or 1, 1, 0
That is 3C1 or 3C2 * 2 ! (3C1 is the no. of ways of choosing a pocket to put everything in. )

That gives 150.
D !
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Re: In how many ways can 5 different colored marbles be placed in 3 distin  [#permalink]

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11 Oct 2019, 08:56
P1 can have either 1 marble or 2 marble or 3 marble.

Therefore,

P1 can have 1 marble in 5 ways as all are of different colors.

Or

P1 can have 2 marbles in 5C2 ways (not 5P2 as order is not important; meaning P1 can have m1m2 or m2m1 and it will mean the same thing). 10 ways.

Or

P1 can have 3 marbles in 5C3 ways. 10 ways.

Hence P1 can have marbles in 25 ways.

Every time we select these marbles their arrangement is also important among the 3 pockets. Because if P1 selects M1 first then it will have repercussion on the selection by the next 2 pockets. Similarly if P1 selects M2 first it will again have an effect on the selection by the next 2 pockets. This means that the arrangement among the 3 pockets shall be considered.

So now we will arrange these 25 number of ways among the 3 pockets.

Therefore,

25 x 3! = 25 X 6 = 150.

Kindly correct if there is anything wrong with my logic. Thanks.

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Re: In how many ways can 5 different colored marbles be placed in 3 distin   [#permalink] 11 Oct 2019, 08:56
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Q&A

Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$.

+3
−0

Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$.

This was a simple observation I made when playing around and I came up with the following proof:

It follows from $(10a + 25)^2 = 100a^2 + 500a + 625 = 100(a^2 + 5a) + 625$ that any number ending in $25$ raised to a power of $2$ will also end in $25$. As each number has a unique binary representation, every number can be written as a sum of powers of two. By the rule $a^{b + c} = a^b \cdot a^c$ and $ab \bmod {c} = (a \bmod {c}) \cdot (b \bmod {c}) \bmod {c}$, $25^n \mod {100}$ will eventually cascade down to $25 \bmod {100}$. For example, \begin{align}25^{19} \equiv 25^{16 + 2 + 1} \equiv 25^{16} \cdot 25^{2} \cdot 25^{1} \equiv 25 \cdot 25 \cdot 25 \\ \equiv 25^{3} \equiv 25^{2 + 1} \equiv 25^2 \cdot 25^1 \equiv 25 \cdot 25 \equiv 25^2 \equiv 25\bmod{100}\end{align}

I am wondering if there are alternate, potentially simpler proofs.

Why does this post require moderator attention?
Why should this post be closed?

+4
−0

There is a fairly simple proof by induction.

Base case: $n = 1$

$$25^1\equiv 25 \mod 100$$

Inductive case: Assuming $25^n\equiv 25 \mod 100$,

\begin{align} 25^{n+1}&\equiv 25\cdot 25^n \mod 100\\ &\equiv 25 \cdot 25 \\ &\equiv 625 \\ &\equiv 25 \end{align}

Actually, this proof can easily be extended to show the stronger statement that $5^n\equiv 25\mod 100$ for all natural numbers $n\ge2$

Base case: $n=2$

$$5^2\equiv25\mod100$$

Inductive case: Assuming $5^n\equiv 25 \mod 100$,

\begin{align} 5^{n+1}&\equiv 5\cdot 5^n \mod 100\\ &\equiv 5 \cdot 25 \\ &\equiv 125 \\ &\equiv 25 \end{align}

The case with powers of $25$ then simply comes from the case when $n$ is even.

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+3
−0

You can prove it using the binomial theorem.

Assume that $1\leq n∈\mathbb{N}$, then:

\begin{align} 25^n & =(20+5)^n=\sum_{k=0}^n\binom{n}{k}20^k\cdot 5^{n-k}=5+\sum_{k=1}^{n-1}\binom{n}{k}20^k\cdot 5^{n-k}+20\\ & =25+5\cdot 20\cdot\sum_{k=1}^{n-1}\binom{n}{k}20^{k-1}\cdot 5^{n-k-1}\equiv 25\mod 100 \end{align}

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## Linear Algebra and Its Applications, Exercise 3.3.6

Exercise 3.3.6. Given the matrix $A$ and vector $b$ defined as follows

$A = \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \qquad b = \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix}$

find the projection of $b$ onto the column space of $A$.

Decompose the vector $b$ into the sum $p + q$ of two orthogonal vectors $p$ and $q$ where $p$ is in the column space. Which subspace is $q$ in?

Answer: We have $p = A(A^TA)^{-1}A^Tb$ per equation (3) of 3L on page 156. We first compute

$A^TA = \begin{bmatrix} 1&2&-2 \\ 1&-1&4 \end{bmatrix} \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix}$

$= \begin{bmatrix} 9&-9 \\ -9&18 \end{bmatrix}$

and then compute

$(A^TA)^{-1} = \frac{1}{9 \cdot 18 - (-9)(-9)} \begin{bmatrix} 18&-(-9) \\ -(-9)&9 \end{bmatrix}$

$= \frac{1}{81} \begin{bmatrix} 18&9 \\ 9&9 \end{bmatrix} = \begin{bmatrix} \frac{2}{9}&\frac{1}{9} \\ \frac{1}{9}&\frac{1}{9} \end{bmatrix}$

Finally we compute

$p = A(A^TA)^{-1}A^Tb$

$= \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \begin{bmatrix} \frac{2}{9}&\frac{1}{9} \\ \frac{1}{9}&\frac{1}{9} \end{bmatrix} \begin{bmatrix} 1&2&-2 \\ 1&-1&4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix}$

$= \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \begin{bmatrix} \frac{2}{9}&\frac{1}{9} \\ \frac{1}{9}&\frac{1}{9} \end{bmatrix} \begin{bmatrix} -9 \\ 27 \end{bmatrix}$

$= \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 6 \end{bmatrix}$

Since $b = p + q$ we have

$q = b - p = \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix} - \begin{bmatrix} 3 \\ 0 \\ 6 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}$

We have

$p \cdot q = 3 \cdot (-2) + 0 \cdot 2 + 6 \cdot 1 = -6 + 0 + 6 = 0$

so that $p$ and $q$ are orthogonal.

The vector $p$ is in $\cal{R}(A)$, the column space of A, and the orthogonal subspace of $\cal{R}(A)$ is $\cal{N}(A^T)$, the left nullspace of $A$. Since $p$ is in $\cal{R}(A)$ and $q$ is orthogonal to $p$, $q$ must be in $\cal{N}(A^T)$, so that $A^Tq = 0$. We confirm this:

$A^Tq = \begin{bmatrix} 1&2&-2 \\ 1&-1&4 \end{bmatrix} \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}$

$= \begin{bmatrix} -2 + 4 -2 \\ -2 - 2 + 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0$

UPDATE: I corrected the calculation of $q$; thanks go to KTL for pointing out the error.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

This entry was posted in linear algebra and tagged , , , , . Bookmark the permalink.

### 6 Responses to Linear Algebra and Its Applications, Exercise 3.3.6

1. Teresa says:

Thank you for the great work! It’s been really helpful. I wanted to mention that A transpose by A is [6 -8; -8 18] and not [9 -9; -9 18] that changes the results a bit.

2. hecker says:

I’m glad you find these posts useful. I’m sorry it’s been a long time since I published the last one.

I don’t understand your comment. Are you referring to A transpose multiplied by A (on the right)? If so, the (1, 1) element of the result matrix is 1 x 1 + 2 x 2 + (-2) x (-2) = 1 + 4 + 4 = 9. I don’t know how you got 6 for that result. Similarly the (1, 2) entry is 1 x 1 + 2 x (-1) + (-2) x 4 = 1 – 2 – 8 = -9 (not -8) and the (2, 1) entry is 1 x 1 + (-1) x 2 + 4 x (-2) = 1 – 2 – 8 = -9 again.

• KTL says:

The question in the 4th edition is different. Hence the confusion for the person above 🙂

• hecker says:

Ah, thanks for the explanation. I’ve never looked at a copy of the 4th edition so I don’t know how the exercises differ.

3. KTL says:

q you have given as p-b though the calculation is done as b-p (correctly)

• hecker says:

Thanks for finding this error! I’ve corrected the post.


# Number of true relations of the form $A\subseteq B$ where $A,B\in\mathcal{P}(\{1,2,\ldots,n\})$

I just started "Introduction to Topology and Modern Analysis" by G.F. Simmons and came across this problem in the exercises.

Q. Let $U=\{1,2,\ldots,n\}$ for an arbitrary positive integer $n$. If $A$ and $B$ are arbitrary subsets of $U$, how many relations of the form $A\subseteq B$ are there? How many of them are true?

Some previous problems for the cases $n=1,2,3$ suggested that by "relations of the form $A\subseteq B$", they mean even the ones where $A\subseteq B$ isn't true.

It would be obvious that there are $2^{2n}=4^n$ such relations since $U$ has $2^n$ subsets, $A,B$ can each be chosen in $2^n$ ways, so there are $2^n\times 2^n=4^n$ such relations.

Now, I think $3^n$ of these relations are true. Here's my idea/proof:

Let us consider $x,y\in\mathcal{P}(U)$ where $\mathcal{P}(U)$ denotes the power set of $U$. Let $x$ have $k$ elements from $U$ (where $0\leq k\leq n$) with $k=0$ corresponding to $x=\emptyset$.

For $x\subseteq y$ to be true, $y$ must have all the elements of $x$ and may/may not have elements from $U\setminus x$. We can construct $y$ by taking $m$ additional elements from $U\setminus x$ where $0\leq m\leq n-k$ which can be done in $\binom{n-k}m$ ways for each value of $m$.

So, for each $x$, we can construct $y$ in $\sum\limits_{m=0}^{n-k}\binom{n-k}m=2^{n-k}$ ways.

Now, we can take $x$ in $\binom nk$ ways for each value of $k$ and hence the number of true relations is $\sum\limits_{k=0}^n\binom nk 2^{n-k}=3^n$.

Is the above correct/rigorous enough?

Also, if the above solution is correct, doesn't it work for any arbitrary set $U$ of cardinality $n$ ?

• what do you mean? You want to know with how many relations you can endow the set $\mathcal P\{1,2,3\dots n\}$ ?? And then how many of these are subsets of the relation $\subseteq$ ? – Jorge Fernández Hidalgo Jul 24 '16 at 18:18
• @CarryonSmiling, it's a problem in the book by Simmons that how many of the set inclusion relations $A\subseteq B$ where $A,B\in\mathcal{P}(\{1,2,\ldots,n\})$ are true? I'm looking for proofreading by the community on my work. – analysis123 Jul 24 '16 at 18:28
• Yeah, your solution looks good. In fact your solution was clearer to me that the actual question. – Jorge Fernández Hidalgo Jul 24 '16 at 18:29

## 2 Answers

I finally got it.

The question seems to be: How many pairs $(A,B)$ of subsets of $\{1,2\dots n\}$ exist so that $A\subseteq B$.

For each element $x\in \{1,2\dots n\}$ we have three choices:

• $x$ is only in $B$
• $x$ is in $A$ and $B$
• $x$ is not in $A$ and not in $B$.

Therefore there are $3^n$ possible pairs.

Yes, this is correct, and yes, only the cardinality of $U$ matters.

You could slightly simplify the proof by noting that there are $2^{n-k}$ subsets of $U\setminus x$, so you don't have to sum over binomial coefficients.

I find the combinatorial proof in Carry on Smiling's answer simpler and more elegant.


# Proof of equation with nested, dependent integrals

As part of a larger proof, I am currently trying to prove the following equation: $$\int_0^r \int_0^{r-t_1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \dots t_k^{m_k - 1} dt_k \dots dt_1 = \frac{r^n}{n!} \prod_{i=1}^{k} (m_i -1)!$$

I know that $n = \sum_{i=1}^k m_i$ and moreover that all $m_i$ are natural numbers.

I've already confirmed with Mathematica that this equation holds for $k \in \{1,\dots,5\}$. I however have some difficulties with proving this equation for any $k > 0$. Any hints would be greatly appreciated!

Here's what I've done so far:

In order to get started, I tried to show it for $k = 1$: $$\int_0^r t_1^{m_1 - 1} dt_1 = \left[\frac{1}{m_1} \cdot t_1^{m_1}\right]_{t_1=0}^{t_1=r} = \frac{r^{m_1}}{m_1} = \frac{r^{m_1}}{m_1!} \cdot \frac{m_1!}{m_1} = \frac{r^{m_1}}{m_1!} \cdot (m_1 - 1)!$$

That was pretty straightforward.

But already for $k=2$ I'm getting stuck:

$$\int_0^r \int_0^{r-t_1} t_1^{m_1 - 1}t_2^{m_2 - 1} dt_2dt_1 = \int_0^r t_1^{m_1 - 1}\int_0^{r-t_1} t_2^{m_2 - 1} dt_2dt_1 = \int_0^r t_1^{m_1 - 1}\left[\frac{1}{m_2} \cdot t_2^{m_2}\right]_{t_2=0}^{t_2=r-t_1}dt_1 = \int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot (r-t_1)^{m_2}dt_1$$

Now if I would like to integrate over $t_1$, I should probably use the binomial theorem, which says: $$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$$ Applied to my case, this results in: \begin{align*} &\int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot \sum_{j=0}^{m_2} {{m_2} \choose j} \cdot r^{m_2-j}\cdot (-1)^j \cdot t_1^jdt_1\\ &= \frac{1}{m_2} \cdot \int_0^r \sum_{j=0}^{m_2} (-1)^j \cdot\frac{m_2!}{j!(m_2-j)!} \cdot r^{m_2-j}\cdot t_1^{m_1 + j - 1}dt_1\\ &=\frac{m_2!}{m_2} \cdot \sum_{j=0}^{m_2} \left((-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \int_0^r t_1^{m_1 + j - 1}dt_1\right)\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} \left((-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \left[\frac{1}{m_1 + j} \cdot t_1^{m_1 + j}\right]_{t_1=0}^{t_1=r}\right)\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \cdot \frac{1}{m_1 + j} \cdot r^{m_1 + j}\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2+m_1} \cdot \frac{1}{m_1 + j}\\ &=r^{m_1 + m_2}(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{1}{(m_1 + j)} \end{align*} Now the only thing left to show is that the sum is equal to $\frac{(m_1-1)!}{(m_1+m_2)!}$. I tried the following: \begin{align*} &\sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{1}{(m_1 + j)}\\ &= \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{m_1 \cdots (m_1+j-1)\cdot(m_1+j+1)\cdots(m_1+m_2)}{m_1 \cdots (m_1 + m_2)} \cdot \frac{(m_1-1)!}{(m_1-1)!}\\ &= \frac{(m_1-1)!}{(m_1+m_2)!} \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot \left(\prod_{i=1, i \neq j}^{m_2} (m_1 + i)\right) \end{align*} So now I only need to show that the remaining sum is equal to one, but I'm not able to do so. Again, playing around with some examples ($m_2 \in \{1,2,3\}$) everything works out. I've tried to reframe it as $\frac{(m_1-1)!}{(m_1+m_2)!m_2!} \cdot \sum_{j=0}^{m_2} (-1)^j \cdot{m_2 \choose j}\cdot \left(\prod_{i=1, i \neq j}^{m_2} (m_1 + i)\right)$, but not with much success.

I'm not quite sure whether I'm on the right track here or whether there's a different approach to this whole thing that looks more promising than what I'm trying to do. I feel like I won't be able to prove the overall thing for $k>0$ if I'm not even able to show it for the special case $k=2$.

I've been working on this for a couple of days now and I don't seem to make progress. If you have any ideas what I could try, that would be great. I'd be very grateful for both help with the $k=2$ proof and for ideas about how to tackle the general $k>0$ proof. Thanks in advance!

Here is a result for $k=2$ (EDIT) and for $k=3$ which shows the general way to proceed.

You are looking at imcomplete beta functions. For $k=2$, I follow your result (calling that $I_2$):

$$I_2 = \int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot (r-t_1)^{m_2}{\rm d}t_1$$

Substituting $t_1 = r z$ gives

$$I_2 = \frac{r^{m_1+m_2}}{m_2} \int_0^1 z^{m_1 - 1}\cdot (1-z)^{m_2} {\rm d} z$$

Now use the definition for the incomplete beta function $B(\cdot,\cdot)$ and its special values for natural numbers:

$$I_2 = \frac{r^{m_1+m_2}}{m_2} B(m_1, m_2+1) = \frac{r^{m_1+m_2}}{m_2} \frac{(m_1 - 1)! m_2 !}{(m_1+m_2)!} = r^{m_1+m_2} \frac{(m_1 - 1)! (m_2 -1) !}{(m_1+m_2)!}$$

which is the desired result for $k=2$.

One step further, for $k=3$. The $r$'s have been substituted away as above.

$$I_3 = \frac{r^n}{m_3} \int_0^1 z_1^{m_1 - 1} \int_0^{1-z_1} z_2^{m_2 - 1}\cdot (1-z_1-z_2)^{m_3} {\rm d}z_2{\rm d}z_1$$

Now for the inner integral, substitute $z_2 = s_2 (1-z_1)$ which gives $$I_3 = \frac{r^n}{m_3} \int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} \int_0^{1} s_2^{m_2 - 1}\cdot (1-s_2)^{m_3} {\rm d}s_2{\rm d}z_1$$

This is a formulation where the inner integral has been made independent on the outer one - and this will work out for larger $k$ as well!

Making this explicit, we get $$I_3 = \frac{r^n}{m_3} \Big[\int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} {\rm d}z_1 \Big] \cdot \Big[\int_0^{1} s_2^{m_2 - 1}\cdot (1-s_2)^{m_3} {\rm d}s_2\Big]$$

Now using the already known results for the Beta function above, $$I_3 = \frac{r^n}{m_3} \frac{(m_1 - 1)! (m_2 + m_3) !}{(m_1+m_2+m_3)!} \cdot \frac{(m_2 - 1)! m_3 !}{(m_2+m_3)!} = {r^n} \frac{(m_1 - 1)! (m_2 - 1)! (m_3 -1)!}{(m_1+m_2+m_3)!}$$

as desired.

From here, you can go on to larger $k$.

• Thanks a lot, the variable change and the Beta function really do the trick! I'll try to formulate the proof for general $k$ but it really looks like it's not as difficult as I thought. Apr 7 '17 at 9:42

So building on the solution of @Andreas for $k=2$ and $k=3$, I was able to prove the whole thing for arbitrary $k>0$:

First of all, one can observe that for any natural number $j > 0$ the following holds: $$\int_0^{r-\sum_{i=1}^{j-1}t_i} t_j^{a-1} \cdot \left(r- \sum_{i=1}^j t_i\right)^b dt_j = B(a,b+1) \cdot \left(r- \sum_{i=1}^{j-1} t_i\right)^{a+b}$$ We can define $t_j = \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot z$ which gives $dt_j = \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot dz$. Making a variable change in the left part of the equation results in $$\int_0^{1} \left(r- \sum_{i=1}^{j-1} t_i\right)^{a-1} \cdot z^{a-1} \cdot \left(r- \sum_{i=1}^{j-1} t_i - \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot z\right)^b \cdot\left(r- \sum_{i=1}^{j-1} t_i\right) dz$$ $$= \left(r- \sum_{i=1}^{j-1} t_i\right)^{a-1+b+1} \int_0^{1} z^{a-1} (1-z)^b dz = B(a,b+1) \cdot \left(r- \sum_{i=1}^{j-1} t_i\right)^{a+b}$$

Now we look at the overall problem: $$I_k = \int_0^r \int_0^{r-t_1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \dots t_k^{m_k - 1} dt_k \dots dt_1$$ This can be rewritten as $$I_k = \int_0^r t_1^{m_1 - 1} \int_0^{r-t_1} t_2^{m_2 - 1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_k^{m_k - 1} dt_k \dots dt_1$$

Using the observation made above, we can solve the innermost integral by setting $j=k, a= m_k, b=0$ which gives us $B(m_k,1) \cdot \left(r- \sum_{i=1}^{k-1} t_i\right)^{m_k}$

We thus get: $$I_k = B(m_k,1) \cdot \int_0^r t_1^{m_1 - 1} \dots \int_0^{r-\sum_{i=1}^{k-2} t_i} t_{k-1}^{m_{k-1} - 1} \cdot \left(r- \sum_{i=1}^{k-1} t_i\right)^{m_k} dt_k \dots dt_1$$

Again, we apply our observation to the innermost integral ($j=k-1,a=m_{k-1},b={m_k}$, resulting in $B(m_{k-1},m_k+1) \cdot \left(r- \sum_{i=1}^{k-2} t_i\right)^{m_{k-1}+m_k}$

Recursively applying this step finally results in: $$I_k = B(m_k,1)\cdot B(m_{k-1},m_k + 1) \cdot B(m_{k-2},m_{k-1}+m_k + 1)\cdot \dots \cdot B(m_1, m_2+\dots+m_k+1)\cdot r^{m_1+\dots+m_k}$$ As all $m_i$ are natural numbers, we can use that $B(x,y)=\frac{(x-1)!\,(y-1)!}{(x+y-1)!}$: $$I_k = r^{m_1+\dots+m_k} \cdot \frac{(m_k-1)!\,0!}{m_k!} \cdot \frac{(m_{k-1}-1)!\,m_k!}{(m_{k-1}+m_k)!} \cdot \frac{(m_{k-2}-1)!\,(m_{k-1}+m_k)!}{(m_{k-2}+m_{k-1}+m_k)!} \cdot \dots \cdot \frac{(m_1-1)!\,(m_2+\dots+m_k)!}{(m_1+m_2+\dots+m_k)!}$$ This reduces to: $$I_k = r^{m_1+\dots+m_k} \cdot (m_k -1)! \cdot \dots \cdot (m_1 -1)! \cdot \frac{1}{(m_1+\dots+m_k)!}$$ With $m_1+\dots+m_k = n$, we can rewrite this as: $$I_k = \frac{r^n}{n!} \prod_{i=1}^{k} (m_i -1)!$$

• Nice! That's the general way to proceed. Apr 7 '17 at 13:38


# Moderate Percentages Solved QuestionAptitude Discussion

 Q. If the price of petrol increases by $25%$, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?
 ✖ A. $25\%$ ✖ B. $16.67\%$ ✔ C. $20\%$ ✖ D. $33.33\%$

Solution:
Option(C) is correct

Let the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol. Therefore, his expense on petrol = $100\times 1$ = Rs.100

Now, the price of petrol increases by $25\%$. Therefore, the new price of petrol = Rs.125.

As he has to maintain his expenditure on petrol constant, he will be spending only Rs.100 on petrol.

Let ‘$x$’ be the number of litres of petrol he will use at the new price.

Therefore,

$125\times x=100$

$\Rightarrow x=\dfrac{100}{125}$

$=\dfrac{4}{5}=0.8$litres

He has cut down his petrol consumption by 0.2 litres

$=\dfrac{0.2}{1}\times 100$

$=20\%$ reduction.

There is a shortcut for solving this problem.

If the price of petrol has increased by $25\%$, it has gone up $1/4^{th}$ of its earlier price.

Therefore, the $\%$ of reduction in petrol that will maintain the amount of money spent on petrol constant

$=\dfrac{1}{4+1}$

$=\dfrac{1}{5}=20\%$

i.e. Express the percentage as a fraction. Then add the numerator of the fraction to the denominator to obtain a new fraction. Convert it to percentage - that is the answer.

Edit: For direct formula involving such questions, check comment by Rushi and Chirag Goyal.

Edit: For more insights on the question, check comment by Priya.

## (5) Comment(s)

Vijay
()

(R*100)/(100+R)..... this formula we will use here.

25*100/(125)=20

so 20% is anwser.

Param
()

Simplest of all . No Need to remember formula. No advance calculation. Based on basic calculation.

Let assume original price be 100. Increased by 25% become Rs125.

Logic : Now we have to keep the consumption same,so we have to bring Rs 125 back to Rs 100. How much we have to decrease so that Rs 125 become Rs 100.

(125-100)125*100 % = 20%

Done . No need to go to consumption. You can calculate using one variable only.

Priya
()

25% means 1/4 and 20% means 1/5. If the question is given as "increased by 25" then $\dfrac{1}{(4+1)}=\dfrac{1}{5}=20\%$

If the question is given as "decreased by 25" then $\dfrac{1}{(4-1)}=\dfrac{1}{3}=33\dfrac{1}{3}\%$

i.e., the value of $33\dfrac{1}{3}\%$ is $\dfrac{1}{3}$

It is easy if you know the values for all the percentages!

Rushi
()

There is shortcut formula for this kind of sums which is $\left(\dfrac{R}{100+R}\right) \times 100$ in the case of rising and $\left(\dfrac{R}{100-R}\right) \times 100$ in the case of fall.

But this formula is applicable only when exp. have to remain constant.

Chirag Goyal
()

Thanks $Rushi$

I'm Rewriting the same Shortcut using TeX commands for better understanding.

$\text{Consumption Should be Decrease or Increase}$

$\text=\dfrac{R}{100\pm R}\times{100}$


# Prove subset of $\mathbb R^2$ is open

Problem. Let $G = \{(x,y): x \ne y\}$. Prove $G$ is an open subset of $\mathbb R^2$.

What I am thinking: If I could show that $\mathbb R^2 \setminus G = \{(x,y): x = y\}$ is a closed set, then its complement $G$ is open. I might be totally off. Any suggestions?

• How did you define an open/closed set? – Listing Nov 24 '13 at 20:38

Let the function $f\colon \mathbb R^2\rightarrow \mathbb R,$ $(x,y)\mapsto x-y$ then $f$ is continuous since it's a polynomial on $x$ and $y$. We have $$G=f^{-1}(\mathbb R^*)$$ Do you know what theorem we should use to conclude?

• Would the theorem be if f is continuous then the inverse image under f of any open set is open? I am only confused on what is meant by R*? – MDW Nov 24 '13 at 20:58
• Yes this is the needed theorem and $\mathbb R^*$ is $\mathbb R\setminus \{0\}$. – user63181 Nov 24 '13 at 21:00
• @Sami: Why can't we just show $\Bbb{R}^2 - G$ is open? – Don Larynx Nov 24 '13 at 21:03
• Do you mean show that $\mathbb R^2-G$ is closed? Yes it's possible and I want avoid this since I don't know his definition of closed set @DonLarynx – user63181 Nov 24 '13 at 21:07
• @SamiBenRomdhane: It is open if GOD wants. :D – mrs Nov 25 '13 at 2:40

Just partition $G$ into two disjoint sets: Let $G_1 = \{(x,y) : x > y \}$ and $G_2 = \{(x,y) : x < y \}$, then clearly $G = G_1 \cup G_2$.

Now, WLOG, let's show that $G_1$ is open. Let $\textbf{x} = (x_1,y_1) \in G_1$, and $B_\textbf{x}(r)$ be the ball around $\textbf{x}$ of radius $r$. In order for $G_1$ to be open, we need to find an $r$ such that $B_\textbf{x}(r) \subset G_1$. But, just choose $r$ to be the shortest distance between $x$ and the line $y=x$. Informally, we know the shortest distance between a point to a line is just the intersection of the normal line with the point $\textbf{x}$, and the line y = x (in our case). The normal line will be $y = -x + x_1 + y_1$ (by a quick computation), and if we intersect this line with $y=x$, we get $x = \frac{x_1 + y_1}{2}$, so we can compute the distance to be $\frac{x_1 - y_1}{\sqrt 2}$. choose $r$ to be this, and we have shown $G_1$ will be open.

Do the same for $G_2$, then $G$ is the union of 2 open sets, and we are done.

The definition of an open set : A subset $U$ of a metric space $(M,d)$ is called \textit{open} if, given any point $x$ in $U$, there exists a real number $ε > 04$ such that, given any point $y$ in $M$ with $d(x, y) < ε$, $y$ also belongs to $U$.

In the case of a line in the plane given by the equation $ax + by + c= 0$, where $a, b$ and $c$ are real constants with $a$ and $b$ not both zero, the distance from the line to a point $(x_0,y_0)$ is $$\frac{|ax_0 + by_0 +c|}{\sqrt{a^2 + b^2}}.$$ Now, we are interested in one specific line, namely $y = x$, i.e. $x - y = 0$.

The distance of any point $(x_0, y_0)$ in $G$ to the line $x -y = 0$ is therefore $$r :=\frac{|x_0 -y_0|}{\sqrt{2}}.$$

Now, consider the open ball centered in $(x_0,y_0)$ and of radius $r$. Do every element of this ball belong to $G$ ?

Suggestion (you were looking for):

Use the theorem that says that the graph of a real continuous function is closed.

$A=\{(x,y)\in \mathbb{R}^2 : x=y\}$ is the graph of the continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$. Where $$f(x)=x$$.

Hence A is closed. Therefore, its complement is open.

The theorem does not hold in general but it certainly does for Hausdorff spaces. All metric spaces ($\mathbb{R}$ is a metric space) are Hausdorff.
Theorem: A topological space $X$ is Hausdorff iff its diagonal is closed.
The diagonal is defined as $$\{(x,x):x\in X\}$$
Note: The diagonal is closed in the product topology on $X\times X$.
HINT: $\Bbb{R}^2 - G$ results in a line. Can this complete $G$?


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# PLS HELP

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The first six rows of Pascal's triangle are shown below, beginning with row zero. Except for the $1$ at each end, row $4$ consists of only even numbers, as does row $2.$ How many of the first $20$ rows have this property? (Don't include row $0$ or row $1$). \begin{tabular}{ccccccccccc} &&&&&1&&&&&\\ &&&&1&&1&&&&\\ &&&1&&2&&1&&&\\ &&1&&3&&3&&1&&\\ &1&&4&&6&&4&&1&\\ 1&&5&&10&&10&&5&&1\\ \end{tabular}

Apr 9, 2021

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Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is $$\boxed{4}$$.

My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as $$\frac{n!}{p!(n-p)!}$$, where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.

Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question

Apr 9, 2021

#1
+420
+1

Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is $$\boxed{4}$$.

My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as $$\frac{n!}{p!(n-p)!}$$, where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.

Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question

textot Apr 9, 2021


# Proving that $T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n$ is $\in O(n)$

Show $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n$$ is $$\in O(n)$$.

I will make the bound to be $$\in O(cn)$$ instead.

Proof by strong induction.

• Base case n =1

$$T(1) = c$$ and $$cn=c*1=c$$ $$\checkmark$$

• Inductive Hypothesis : $$T(k) \in O(ck)$$ for $$1\le k1$$.
• Inductive Step: Prove for n. So prove that $$T(n) \le O(ck)$$.

Right away we can write $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n \\ \leq T(n/2)+T(n/4)+T(n/8) + n\\ = c(n/2)+c(n/4)+c(n/8)+n \ \ \ \ By \ Inductive \ Hypothesis$$

$$= c(7n/8)+n \le cn+n ...$$ I am stuck here

*Goal: $$\le cn - (some \ stuff)$$ and some stuff needs to be $$\ge 0$$.

• – ryan Mar 5 at 0:01
• The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Mar 6 at 8:04

First, let me mention that $$O(n)$$ and $$O(cn)$$ are exactly the same thing. What you are really after is showing that $$T(n) \leq cn$$ for all $$n$$.

Let us aim at a slightly more relaxed goal: showing that $$T(n) \leq An$$ for some possibly larger constant $$A$$. For the base case, we need $$A \geq c$$. For the inductive step, we know that $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n \leq A(n/2+n/4+n/8)+1 = (\tfrac{7}{8}A+1)n.$$ Recall that our goal is to deduce that $$T(n) \leq An$$. This would follow if $$\tfrac{7}{8}A + 1 \leq A$$, that is, if $$A \geq 8$$.

In total, if we take $$A = \max(c,8)$$ then both the base case and the inductive step go through.

As an aside, you can also use the Akra-Bazzi theorem to directly conclude that $$T(n) = O(n)$$.

Now let us try to obtain more insight on the recurrence. Let $$S(n) = 8n - T(n)$$. Then $$S(n) = S(\lfloor n/2 \rfloor) + S(\lfloor n/4 \rfloor) + S(\lfloor n/8 \rfloor) + 8n - 8\lfloor n/2 \rfloor - 8\lfloor n/4 \rfloor - 8\lfloor n/8 \rfloor - n.$$ Since $$8(n/a-1) < 8\lfloor n/a \rfloor \leq 8(n/a)$$, we see that $$S(n) = S(\lfloor n/2 \rfloor) + S(\lfloor n/4 \rfloor) + S(\lfloor n/8 \rfloor) + r(n),$$ where $$0 \leq r(n) < 24$$. Applying the Akra-Bazzi theorem, we get $$S(n) = O(n^p)$$ for some $$p < 1$$ (the solution to $$(1/2)^p + (1/4)^p + (1/8)^p = 1$$), and so $$T(n) = 8n + O(n^p)$$.

• Thank you this is an amazing answer and helps a lot! – Mandy Mar 5 at 23:45
• So for base case we can choose either 8 or c (in this case c) , and for inductive step we can also choose 8 or c (in this case 8) ? – Mandy Mar 7 at 20:34
• You have to choose the same constant in both cases. That’s how induction works. – Yuval Filmus Mar 7 at 20:50

It's very important that you understand what f(n) = O (g(n)) means. It means that there is a number $$n_0 ≥ 0$$ and a number c > 0 such that for every $$n ≥ n_0$$, f(n) ≤ c * g(n). It is a property of the whole function, not a property of some n. Saying "I prove by induction that for every n, f(n) = O (g(n))" doesn't make any sense at all. What makes sense is to say "I prove by induction that for every n ≥ n_0, f(n) ≤ c * g(n)".

So what you want to prove per induction using some suitable c, T(n) ≤ cn. One variant of complete induction uses an induction step where you proof "if the statement $$S_k$$ is true for every k < n, then $$S_n$$ is also true".

If T(k) ≤ ck is true for every k < n, then T(n) = T(n/2) + T(n/4) + T(n/8) + n ≤ cn/2 + cn/4 + cn/8 + n = (7/8 c + 1) n.

If c = 1 this means T(n) < 1 7/8 n. Not what we need, we need T(n) < cn. If c = 2 it means T(n) + 2 3/4 n. Slightly better but not good enough. For which c is (7/8 c + 1) n ≤ cn, or 7/8 c + 1 ≤ c? That's the case for c ≥ 8. The start of the induction is n=0, and it's easy to show that T(0) = 0.

So you haven't just shown that T(n) = O(n), you have shown the much stronger T(n) ≤ 8n.

• Thank you this is an excellent answer and helps me understand these kinds of problems in more depth! – Mandy Mar 5 at 23:46
• You might have forgotten the base case. – Yuval Filmus Mar 6 at 16:38


### NBHM 2020 PART A Question 10 Solution: Possible rank of a matrix obtained by changing the entries $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$

Let $M$ be a $7×6$ real matrix. The entries of $M$ in the positions $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$ are changed to obtain another $7×6$ real matrix $\widetilde{M}$. Suppose that the rank of $\widetilde{M}$ is 4. What could be the rank of $M$? List all possibilities.
Solution:
First, we will derive a connection between the rank of $M$ and the rank of $\widetilde{M}$. We have,
$$rank(M) = rank(M-\widetilde{M}+\widetilde{M}) \le rank(M-\widetilde{M})+rank(M)$$ Similarly
$$rank(\widetilde M) = rank(\widetilde{M}-M+M) \le rank(\widetilde{M}-M)+rank(M)$$
This shows that $rank(\widetilde{M}) - rank(\widetilde{M}-M) \le rank(M) \le rank(M-\widetilde{M})+rank(\widetilde{M})$
The matrices $M - \widetilde{M}$ and $\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\widetilde{M}$ is equal to $4$. Hence we have $$2 \le rank(M) \le 6.$$
Using $0,1$ matrics one can check that all these possibilities are occurring indeed.
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