Here we need to talk about cardinality of a set, which is basically the size of the set. there are $10$ people with white shirts and $8$ people with red shirts; $4$ people have black shoes and white shirts; $3$ people have black shoes and red shirts; the total number of people with white or red shirts or black shoes is $21$. that the cardinality of a set is the number of elements it contains. Edition 1st Edition. $$|W \cap B|=4$$ In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides of students who play both foot ball and cricket = 25, No. In Section 5.1, we defined the cardinality of a finite set $$A$$, denoted by card($$A$$), to be the number of elements in the set $$A$$. Total number of elements related to both B & C. Total number of elements related to both (B & C) only. Two sets are equal if and only if they have precisely the same elements. For example, let A  =  { -2, 0, 3, 7, 9, 11, 13 }, Here, n(A) stands for cardinality of the set A. To prove there exists a bijection between to sets X and Y, there are 2 ways: 1. find an explicit bijection between the two sets and prove it is bijective (prove it is injective and surjective) 2. Pages 5. eBook ISBN 9780429324819. It turns out we need to distinguish between two types of infinite sets, Any superset of an uncountable set is uncountable. then talk about infinite sets. The cardinality of a set is roughly the number of elements in a set. (Hint: Use a standard calculus function to establish a bijection with R.) 2. Math 127: In nite Cardinality Mary Radcli e 1 De nitions Recall that when we de ned niteness, we used the notion of bijection to de ne the size of a nite set. of students who play cricket only = 10, No. where one type is significantly "larger" than the other. The cardinality of the set of all natural numbers is denoted by . $$\>\>\>\>\>\>\>+\sum_{i < j < k}\left|A_i\cap A_j\cap A_k\right|-\ \cdots\ + \left(-1\right)^{n+1} \left|A_1\cap\cdots\cap A_n\right|.$$, $= |W| + |R| + |B|- |W \cap R| - |W \cap B| - |R \cap B| + |W \cap R \cap B|$. ... to make the argument more concrete, here we provide some useful results that help us prove if a set is countable or not. Hence these sets have the same cardinality. The cardinality of a set is denoted by $|A|$. of students who play foot ball only = 28, No. $$|W|=10$$ The intuition behind this theorem is the following: If a set is countable, then any "smaller" set $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$, and any of their subsets are countable. c) $(0,\infty)$, $\R$ d) $(0,1)$, $\R$ Ex 4.7.4 Show that $\Q$ is countably infinite. If $B \subset A$ and $A$ is countable, by the first part of the theorem $B$ is also a countable the idea of comparing the cardinality of sets based on the nature of functions that can be possibly de ned from one set to another. Ex 4.7.3 Show that the following sets of real numbers have the same cardinality: a) $(0,1)$, $(1, \infty)$ b) $(1,\infty)$, $(0,\infty)$. countable, we can write a proof, we can argue in the following way. So, the total number of students in the group is 100. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,…,n-1} to the set. If you are less interested in proofs, you may decide to skip them. The set of all real numbers in the interval (0;1). Before we start developing theorems, let’s get some examples working with the de nition of nite sets. In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. In mathematics, a set is a well-defined collection of distinct elements or members. Two finite sets are considered to be of the same size if they have equal numbers of elements. then by removing the elements in the list that are not in $B$, we can obtain a list for $B$, In addition, we say that the empty set has cardinality 0 (or cardinal number 0), and we write $$\text{card}(\emptyset) = 0$$. As far as applied probability Discrete Mathematics and Its Applications, Seventh Edition answers to Chapter 2 - Section 2.5 - Cardinality of Sets - Exercises - Page 176 12 including work step by step written by community members like you. $$|W \cup B \cup R|=21.$$ set whose elements are obtained by multiplying each element of Z by k.) The function f : N !Z de ned by f(n) = ( 1)nbn=2cis a 1-1 corre-spondence between the set of natural numbers and the set of integers (prove it!). The Math Sorcerer 19,653 views. If $A$ has only a finite number of elements, its cardinality is simply the Here is a simple guideline for deciding whether a set is countable or not. the inclusion-exclusion principle we obtain. and how to prove set S is a infinity set. you can never provide a list in the form of $\{a_1, a_2, a_3,\cdots\}$ that contains all the $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$. \mathbb {R}. thus $B$ is countable. Examples of Sets with Equal Cardinalities. This will come in handy, when we consider the cardinality of infinite sets in the next section. 4 CHAPTER 7. Cardinality of a set S, denoted by |S|, is the number of elements of the set. Since each $A_i$ is countable we can Definition. Cantor introduced a new de・］ition for the 窶徭ize窶・of a set which we call cardinality. Textbook Authors: Rosen, Kenneth, ISBN-10: 0073383090, ISBN-13: 978-0-07338-309-5, Publisher: McGraw-Hill Education The difference between the two types is We will say that any sets A and B have the same cardinality, and write jAj= jBj, if A and B can be put into 1-1 correspondence. but "bigger" sets such as $\mathbb{R}$ are called uncountable. You already know how to take the induction step because you know how the case of two sets behaves. \mathbb {N} To formulate this notion of size without reference to the natural numbers, one might declare two finite sets A A A and B B B to have the same cardinality if and only if there exists a bijection A → B A \to B A → B. Cardinality Recall (from our first lecture!) (useful to prove a set is finite) • A set is infinite when there is an injection, f:AÆA, such that f(A) is … When an invertible function from a set to \Z_n where m\in\N is given the cardinality of the set immediately follows from the definition. This important fact is commonly known ... aged to prove that two very different sets are actually the same size—even though we don’t know exactly how big either one is. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,…,n-1} to the set. Cardinality The cardinality of a set is roughly the number of elements in a set. By Gove Effinger, Gary L. Mullen. (a) Let S and T be sets. Furthermore, we designate the cardinality of countably infinite sets as ℵ0 ("aleph null"). 12:14. Here we need to talk about cardinality of a set, which is basically the size of the set. In particular, one type is called countable, If A and B are disjoint sets, n(A n B)  =  0, n(A u B u C)  =  n(A) + n(B) + n(C) - n(A n B) - n(B n C)                                  - n(A n C) + n(A n B n C), n(A n B)  = 0, n(B n C)  =  0, n(A n C)  =  0, n(A n B n C)  =  0, = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC). Cardinality of a set: Discrete Math: Nov 17, 2019: Proving the Cardinality of 2 finite sets: Discrete Math: Feb 16, 2017: Cardinality of a total order on an infinite set: Advanced Math Topics: Jan 18, 2017: cardinality of a set: Discrete Math: Jun 1, 2016 One important type of cardinality is called “countably infinite.” A set A is considered to be countably infinite if a bijection exists between A and the natural numbers ℕ. Countably infinite sets are said to have a cardinality of א o (pronounced “aleph naught”). $$|A \cup B |=|A|+|B|-|A \cap B|.$$ Solution. Before discussing thus by subtracting it from $|A|+|B|$, we obtain the number of elements in $|A \cup B |$, (you can The above arguments can be repeated for any set $C$ in the form of However, to make the argument In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. (Assume that each student in the group plays at least one game). Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively. We first discuss cardinality for finite sets and (Hint: you can arrange $\Q^+$ in a sequence; use this to arrange $\Q$ into a sequence.) set which is a contradiction. of students who play both (foot ball & hockey) only = 12, No. To see this, note that when we add $|A|$ and $|B|$, we are counting the elements in $|A \cap B|$ twice, Because of the symmetyofthissituation,wesaythatA and B can be put into 1-1 correspondence. That is often difficult, however. If $A$ is a finite set, then $|B|\leq |A| < \infty$, Then,byPropositionsF12andF13intheFunctions section,fis invertible andf−1is a 1-1 correspondence fromBtoA. Consider sets A and B.By a transformation or a mapping from A to B we mean any subset T of the Cartesian product A×B that satisfies the following condition: . Question: Prove that N(all natural numbers) and Z(all integers) have the same cardinality. To prove that a given in nite set X … Theorem. How to prove that all maximal independent sets of a matroid have the same cardinality. We can, however, try to match up the elements of two inﬁnite sets A and B one by one. As seen, the symbol for the cardinality of a set resembles the absolute value symbol — a variable sandwiched between two vertical lines. Provided a matroid is a 2-tuple (M,J ) where M is a finite set and J is a family of some of the subsets of M satisfying the following properties: If A is subset of B and B belongs to J , then A belongs to J , =  n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC), n(FuHuC)  =  65 + 45 + 42 -20 - 25 - 15 + 8. This function is bijective. When it ... prove the corollary one only has to observe that a function with a “right inverse” is the “left inverse” of that function and vice versa. No. Total number of students in the group is n(FuHuC). $\mathbb{Z}=\{0,1,-1,2,-2,3,-3,\cdots\}$. In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. This establishes a one-to-one correspondence between the set of primes and the set of natural numbers, so they have the same cardinality. A = \left\ { {1,2,3,4,5} \right\}, \Rightarrow \left| A \right| = 5. … uncountable set (to prove uncountability). To this final end, I will apply the Cantor-Bernstein Theorem: (The two sets (0, 1) and [0, 1] have the same cardinality if we can find 1-1 mappings from (0, 1) to [0, 1] and vice versa.) Also known as the cardinality, the number of disti n ct elements within a set provides a foundational jump-off point for further, richer analysis of a given set. The examples are clear, except for perhaps the last row, which highlights the fact that only unique elements within a set contribute to the cardinality. I presume you have sent this A2A to me following the most recent instalment of our ongoing debate regarding the ontological nature and resultant enumeration of Zero. To be precise, here is the definition. Find the total number of students in the group. A set A is countably infinite if and only if set A has the same cardinality as N (the natural numbers). The proof of this theorem is very similar to the previous theorem. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. Now that we know about functions and bijections, we can define this concept more formally and more rigorously. The idea is exactly the same as before. f:A → Bbea1-1correspondence. while the other is called uncountable. However, I am stuck in proving it since there are more than one "1", "01" = "1", same as other numbers. Fix m 2N. $$\biggl|\bigcup_{i=1}^n A_i\biggr|=\sum_{i=1}^n\left|A_i\right|-\sum_{i < j}\left|A_i\cap A_j\right|$$ but you cannot list the elements in an uncountable set. Imprint CRC Press. 1. if it is a finite set, $\mid A \mid < \infty$; or. I could not prove that cardinality is well defined, i.e. Thus according to Deﬁnition 2.3.1, the sets N and Z have the same cardinality. When A and B have the same cardinality, we write jAj= jBj. Thus, any set in this form is countable. of students who play both (hockey & cricket) only = 7, No. A set is an infinite set provided that it is not a finite set. of students who play hockey only = 18, No. The two sets A = {1,2,3} and B = {a,b,c} thus have the cardinality since we can match up the elements of the two sets in such a way that each element in each set is matched with exactly one element in the other set. For example, if $A=\{2,4,6,8,10\}$, then $|A|=5$. Math 131 Fall 2018 092118 Cardinality - Duration: 47:53. For in nite sets, this strategy doesn’t quite work. Let X m = fq 2Q j0 q 1; and mq 2Zg. For example, a consequence of this is that the set of rational numbers $\mathbb{Q}$ is countable. Since $A$ and $B$ are Cardinality of a set is a measure of the number of elements in the set. If A can be put into 1-1 correspondence with a subset of B (that is, there is a 1-1 case the set is said to be countably infinite. I can tell that two sets have the same number of elements by trying to pair the elements up. (useful to prove a set is finite) • A set is infinite when there … (b) A set S is finite if it is empty, or if there is a bijection for some integer . of students who play both foot ball &  hockey = 20, No. n(AuB)  =  Total number of elements related to any of the two events A & B. n(AuBuC)  =  Total number of elements related to any of the three events A, B & C. n(A)  =  Total number of elements related to  A. n(B)  =  Total number of elements related to  B. n(C)  =  Total number of elements related to  C. Total number of elements related to A only. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. We first discuss cardinality for finite sets and then talk about infinite sets. A useful application of cardinality is the following result. It would be a good exercise for you to try to prove this to yourself now. If S is a set, we denote its cardinality by |S|. Cantor showed that not all in・］ite sets are created equal 窶・his de・］ition allows us to distinguish betweencountable and uncountable in・］ite sets. Set S is a set consisting of all string of one or more a or b such as "a, b, ab, ba, abb, bba..." and how to prove set S is a infinity set. For finite sets, cardinalities are natural numbers: |{1, 2, 3}| = 3 |{100, 200}| = 2 For infinite sets, we introduced infinite cardinals to denote the size of sets: $$C=\bigcup_i \bigcup_j \{ a_{ij} \},$$ The cardinality of a set is denoted by $|A|$. Figure 1.13 shows one possible ordering. We have been able to create a list that contains all the elements in $\bigcup_{i} A_i$, so this correspondence with natural numbers $\mathbb{N}$. is concerned, this guideline should be sufficient for most cases. n(FnH)  =  20, n(FnC)  =  25, n(HnC)  =  15. The set whose elements are each and each and every of the subsets is the ability set. The above theorems confirm that sets such as $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and their Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. 11 Cardinality Rules ... two sets, then the sets have the same size. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. For in nite sets, this strategy doesn’t quite work. Any set which is not finite is infinite. If $A_1, A_2,\cdots$ is a list of countable sets, then the set $\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$ Discrete Mathematics - Cardinality 17-16 More Countable Sets (cntd) The fact that you can list the elements of a countably infinite set means that the set can be put in one-to-one If $A$ and $B$ are countable, then $A \times B$ is also countable. should also be countable, so a subset of a countable set should be countable as well. We can say that set A and set B both have a cardinality of 3. set is countable. S and T have the same cardinality if there is a bijection f from S to T. Notation: means that S and T have the same cardinality. respectively. The cardinality of a set is denoted by $|A|$. Thus by applying there'll be 2^3 = 8 elements contained in the ability set. Definition of cardinality. forall s : fset_expr (A:=A), exists n, (cardinality_fset s n /\ forall s' n', eq_fset s s' -> cardinality_fset s' n' -> n' = n). 1. On the other hand, you cannot list the elements in $\mathbb{R}$, DOI link for Cardinality of Sets. so it is an uncountable set. | A | = | N | = ℵ0. Mappings, cardinality. The number of elements in a set is called the cardinality of the set. This poses few diﬃculties with ﬁnite sets, but inﬁnite sets require some care. The sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. $$B = \{b_1, b_2, b_3, \cdots \}.$$ and Itiseasytoseethatanytwoﬁnitesetswiththesamenumberofelementscanbeput into1-1correspondence. For example, you can write. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides The cardinality of a finite set is the number of elements in the set. if you need any other stuff in math, please use our google custom search here. Any subset of a countable set is countable. If you are less interested in proofs, you may decide to skip them. Total number of elements related to C only. refer to Figure 1.16 in Problem 2 to see this pictorially). I can tell that two sets have the same number of elements by trying to pair the elements up. Click here to navigate to parent product. However, as we mentioned, intervals in $\mathbb{R}$ are uncountable. Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. 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# jacobian

Jacobian matrix

## Description

example

jacobian(f,v) computes the Jacobian matrix of f with respect to v. The (i,j) element of the result is $\frac{\partial f\left(i\right)}{\partial \text{v}\left(j\right)}$.

## Examples

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The Jacobian of a vector function is a matrix of the partial derivatives of that function.

Compute the Jacobian matrix of [x*y*z,y^2,x + z] with respect to [x,y,z].

syms x y z
jacobian([x*y*z,y^2,x + z],[x,y,z])
ans =

$\left(\begin{array}{ccc}y z& x z& x y\\ 0& 2 y& 0\\ 1& 0& 1\end{array}\right)$

Now, compute the Jacobian of [x*y*z,y^2,x + z] with respect to [x;y;z].

jacobian([x*y*z,y^2,x + z], [x;y;z])
ans =

$\left(\begin{array}{ccc}y z& x z& x y\\ 0& 2 y& 0\\ 1& 0& 1\end{array}\right)$

The Jacobian matrix is invariant to the orientation of the vector in the second input position.

The Jacobian of a scalar function is the transpose of its gradient.

Compute the Jacobian of 2*x + 3*y + 4*z with respect to [x,y,z].

syms x y z
jacobian(2*x + 3*y + 4*z,[x,y,z])
ans = $\left(\begin{array}{ccc}2& 3& 4\end{array}\right)$

Now, compute the gradient of the same expression.

ans =

$\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)$

The Jacobian of a function with respect to a scalar is the first derivative of that function. For a vector function, the Jacobian with respect to a scalar is a vector of the first derivatives.

Compute the Jacobian of [x^2*y,x*sin(y)] with respect to x.

syms x y
jacobian([x^2*y,x*sin(y)],x)
ans =

$\left(\begin{array}{c}2 x y\\ \mathrm{sin}\left(y\right)\end{array}\right)$

Now, compute the derivatives.

diff([x^2*y,x*sin(y)],x)
ans = $\left(\begin{array}{cc}2 x y& \mathrm{sin}\left(y\right)\end{array}\right)$

Specify polar coordinates $r\left(t\right)$, $\varphi \left(t\right)$, and $\theta \left(t\right)$ that are functions of time.

syms r(t) phi(t) theta(t)

Define the coordinate transformation form spherical coordinates to Cartesian coordinates.

R = [r*sin(phi)*cos(theta), r*sin(phi)*sin(theta), r*cos(phi)]
R(t) = $\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)& \mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) r\left(t\right)\end{array}\right)$

Find the Jacobian of the coordinate change from spherical coordinates to Cartesian coordinates.

jacobian(R,[r,phi,theta])
ans(t) =

$\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) \mathrm{cos}\left(\theta \left(t\right)\right) r\left(t\right)& -\mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)\\ \mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)& \mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)\\ \mathrm{cos}\left(\varphi \left(t\right)\right)& -\mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)& 0\end{array}\right)$

## Input Arguments

collapse all

Scalar or vector function, specified as a symbolic expression, function, or vector. If f is a scalar, then the Jacobian matrix of f is the transposed gradient of f.

Vector of variables or functions with respect to which you compute Jacobian, specified as a symbolic variable, symbolic function, or vector of symbolic variables. If v is a scalar, then the result is equal to the transpose of diff(f,v). If v is an empty symbolic object, such as sym([]), then jacobian returns an empty symbolic object.

collapse all

### Jacobian Matrix

The Jacobian matrix of the vector function f = (f1(x1,...,xn),...,fn(x1,...,xn)) is the matrix of the derivatives of f:

$J\left({x}_{1},\dots {x}_{n}\right)=\left[\begin{array}{ccc}\frac{\partial {f}_{1}}{\partial {x}_{1}}& \cdots & \frac{\partial {f}_{1}}{\partial {x}_{n}}\\ ⋮& \ddots & ⋮\\ \frac{\partial {f}_{n}}{\partial {x}_{1}}& \cdots & \frac{\partial {f}_{n}}{\partial {x}_{n}}\end{array}\right]$

## Version History

Introduced before R2006a


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# In how many different ways can the letters of the word MISSISSIPPI be

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In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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08 Apr 2016, 01:47
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In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?

A. 48
B. 144
C. 210
D. 420
E. 840
[Reveal] Spoiler: OA

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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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08 Apr 2016, 20:21
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The word MISSISSIPPI has 4 I , 4S , 2P and M .
The only vowel present in the word is I and we have to consider 4 I's as a single group

Number of different ways = 8!/(4!*2!)
=(8*7*6*5)/2
=840
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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26 Jun 2017, 11:18
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?

A. 48
B. 144
C. 210
D. 420
E. 840

vowel (I) together can be arranged with other 7 letters in 8! ways
4 no. vowels themselves can be arranged in 4! ways

different ways = 8!*4!/( 4!*4!*2!) = 840---(4I's , 4 S's , 2 P's)

Ans E

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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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26 Jun 2017, 12:18
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?

A. 48
B. 144
C. 210
D. 420
E. 840

Ooh, a combinatorics problem with some fairly large numbers. The best technique for these is to mentally break them down into simpler problems. Otherwise, the solutions tend to sort of look like magic tricks - sure, you can use the formula, but how are you supposed to know to use that formula, and not one of the many similar-looking ones?

There are four vowels, and they're all the same. Let's set those vowels aside: (IIII)

Now we have the letters MSSSSPP. How many ways can just those letters be arranged? Well, if they were all different, we could arrange them in 7x6x5x4x3x2x1 = 7! ways. However, they aren't all different. For instance, four of the letters are (S). I'm going to color them to demonstrate why that matters:

one arrangement: MSSSSPP
a 'different' arrangment: MSSSSPP

We counted both of those arrangements, but we actually don't want to. All of the Ss look the same - they aren't actually different colors - so we want to make those arrangements the same. Because there are 4x3x2x1 = 24 ways to order the different Ss, we want every set of 24 arrangements where the Ss are in the same place, to just count as 1 arrangement, instead. So, we can divide out the extra possibilities by dividing our total by 24 (or 4!):

7!/4!

Do the same thing with the two Ps. We still have twice as many arrangements as we need, since we've counted as if the two Ps were different, but they're actually the same. So, divide the total by 2:

7!/(4! x 2)

Now we have to put the (IIII) letters back in. They all have to go together. Start by looking at one of the arrangements of the other letters: PMSSPSS. Where can the four Is go? There are 8 places where we can put them:

IIIIPMSSPSS
PIIIIMSSPSS
PMIIIISSPSS
PMSIIIISPSS

etc.

So, for each arrangement, we have to multiply by 8, to account for the eight possible ways to put the vowels back in. Here's the final answer:

(8 x 7!) / (4! x 2) = (8 x 7 x 6 x 5 x 4 x 3 x 2) / (4 x 3 x 2 x 2) = 4 x 7 x 6 x 5 = 4 x 210 = 840.
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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26 Jun 2017, 12:40
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?

A. 48
B. 144
C. 210
D. 420
E. 840

When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

-------NOW ONTO THE QUESTION!!!-----------------

First "glue" the 4 I's together to create ONE character: IIII (this ensures that they stay together)
So, basically, we must determine the number of arrangements of M, S, S, S, S, P, P and IIII
There are 8 characters in total
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 8!/[(4!)(2!)]
= 840

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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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04 Dec 2017, 08:31

Stage 1: Lets place the letters without the 4I's:
MSSSSPP
7!/(4!2!)= 105 ways

Stage 2: Now lets place the 4I's
-m-s-s-s-s-p-p-
We can place th4 4I's in 8 ways

Total ways= 105*8= 840 ways.

Is this a correct approach?

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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]

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04 Dec 2017, 09:08
Rearranging letters has a simple formula:

Total number of letters
*DIVIDED BY*
(2)! for every letter repeated twice / (3)! for every letter repeated thrice and so on...

Mississippi has:

4 s's
4 i's
2 p's

Total number of letters: 11!

So the answer is $$\frac{11!}{4!4!2!}$$
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Re: In how many different ways can the letters of the word MISSISSIPPI be   [#permalink] 04 Dec 2017, 09:08
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# In how many different ways can the letters of the word MISSISSIPPI be

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$10$ people ($6$ male, $4$ female) divided into $2$ equal groups: what is the probability that all females are in the same group?

This question comes from a completed, marked, and returned exam. It will not likely be reused.

Problem

As stated in the question above

Work

First, I note that there are $$\binom{10}{5}$$ possible groupings.

Second, I note that, if all $$4$$ females are in the same group, then the remaining fifth member is one of the boys: there are $$\binom{6}{1} = 6$$ ways to choose the fifth member.

So I conclude $$\Pr = \frac{6}{\binom{10}{5}} = \frac{1}{42}$$.

Question

I was marked incorrect: the given answer is $$\frac{1}{21}$$, or exactly twice my answer.

What reasoning led to this conclusion? Why does it seem like some sort of symmetry argument allows us to conclude there are $$12$$ ways to choose the fifth member?

• Note that $\frac{6}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}$. – N. F. Taussig Feb 28 at 16:55
• @N.F.Taussig apologies i was looking at the answers on the question below as i typed – D. Ben Knoble Feb 28 at 17:01
• Another way to see that there are only $126$ possible groups is to observe that if Eloise is one of the four girls, then there are $\binom{9}{4}$ ways to select which four of the other members are in her group. – N. F. Taussig Feb 28 at 17:28

You've double-counted the groupings: $$\{A, B, C, D, E\}$$ and $$\{F, G, H, I, J\}$$ is the same grouping as $$\{F, G, H, I, J\}$$ and $$\{A, B, C, D, E\}$$. Accounting for this double-count, there are $$\frac{1}{2} \binom{10}{5}$$ distinct groupings. (This is probably the most-common counting mistake of all time. Everyone makes it at least once.)


# Show if A,B,C are invertible matrices of same size

1. Jul 7, 2008

Show if A,B,C are invertible matrices of same size....

I know, I know. I should be awesome at these by now...

1. The problem statement, all variables and given/known data
Show if A,B,C are invertible matrices of the same size, than $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$

3. The attempt at a solution Given some matrix A,$AA^{-1}=I$
If:
$AA^{-1}=I$
$BB^{-1}=I$
$CC^{-1}=I$

I am not sure where to go from here. I don't think I have any more definitions or product rules to incorporate.

It almost seems as if I would FIRST have to show that (ABC) is invertible to begin with. Then I can use the fact that (ABC)(ABC)^{-1}=I to discover how (ABC)^{-1} MUST be arranged in order for the product of the two to yield I.

Does that sound like a good place to start? Proving if A,B, and C are invertible, then (ABC) is too?

2. Jul 7, 2008

### rock.freak667

Re: Show if A,B,C are invertible matrices of same size....

Let $D=(ABC)^{-1}$
$\times (ABC)$
$\Rightarrow ABCD=I$

and then just multiply by $A^{-1}$ and so forth

3. Jul 7, 2008

Re: Show if A,B,C are invertible matrices of same size....

I don't follow? What does the second line mean?--->$\times (ABC)$

4. Jul 7, 2008

### rock.freak667

Re: Show if A,B,C are invertible matrices of same size....

Multiply both sides by the matric ABC

5. Jul 7, 2008

Re: Show if A,B,C are invertible matrices of same size....

So
$D=(ABC)^{-1}$
$\Rightarrow D(ABC)=(ABC)^{-1}(ABC)$

Well, I think I see where this is going. And I think that the only reason this works is because we are assuming that the product (ABC) IS invertible. Which brings me back to my original point. In order to show how the multiplication MUST be carried out, we must first SHOW or assume without proof that (ABC) is in fact invertible since our argument will be based on the fact that (ABC)*(ABC)^{-1}=Id.

6. Jul 7, 2008

### rock.freak667

Re: Show if A,B,C are invertible matrices of same size....

Well if A,B,C are nxn matrices then ABC is an nxn matrix.

and for the matrix ABC to be invertible $det(ABC) \neq 0$

and det(ABC)=det(A)*det(B)*det(C)

but the matrices A,B,C are invertible.

7. Jul 7, 2008

### Defennder

Re: Show if A,B,C are invertible matrices of same size....

Another way you could show that a product of two matrices A and B are invertible is by showing that there exists some matrix which when multiplied to AB on the left and on the right gives the identity matrix:

Suppose A and B are invertible, then:

$$AB(B^{-1}A^{-1}) = I$$ for multiplying on the right
$$B^{-1}A^{-1}AB = I$$ for multiplying on the left.

In both cases this reduces to I, so $$B^{-1}A^{-1}$$ is the inverse of AB.

Now make use of this result to prove your question.

8. Jul 7, 2008

### HallsofIvy

Staff Emeritus
Re: Show if A,B,C are invertible matrices of same size....

It is not nnecessary to assume that ABC is invertible.

You are given that A, B, C separately are invertible so A-1, B-1, and C-1 exist. Thus C-1B-1A-1 exists. What do you get if you multiply (ABC)(C-1B-1C-1) and (C-1B-1A-1(ABC)?

That will prove that ABC is invertible.

9. Jul 7, 2008

### matt grime

Re: Show if A,B,C are invertible matrices of same size....

X is invertible if there is a Y with XY=YX=Id.

It isn't that I want to show ABC is invertible, but someone's been really helpful and just asked me to verify what the inverse is! I don't have to think at all, I just have to chuck the nominal inverse into the definition and see what happens. So do it....

10. Jul 7, 2008

### Defennder

Re: Show if A,B,C are invertible matrices of same size....

Hey yeah, I didn't see it that way. Makes it a lot easier.

11. Jul 7, 2008

### JinM

Re: Show if A,B,C are invertible matrices of same size....

This looks like a question that could spring up on my midterm tomorrow. Assuming invertibility, does this prove the result?

$$ABCC^{-1}B^{-1}A^{-1} = A(BI_{n})B^{-1}A^{-1}= A(BB^{-1})A^{-1} = AA^{-1} = I_n$$

Now, the other side,

$$C^{-1}B^{-1}A^{-1}ABC = C^{-1}B^{-1}BC = C^{-1}C = I_n$$

Thus $$(ABC)(C^-1B^-1A^-1}) = (C^-1B^-1A^-1)(ABC) = I_n$$.

Is this enough? No claim of originality, this was how a result was proved in my notes, and what everyone is getting at!

Last edited: Jul 7, 2008
12. Jul 7, 2008

### Defennder

Re: Show if A,B,C are invertible matrices of same size....

That's basically what HallsOfIvy said.

13. Jul 7, 2008

Re: Show if A,B,C are invertible matrices of same size....

I don't like it. I don't know why yet...but I just don't.

14. Jul 7, 2008

### matt grime

Re: Show if A,B,C are invertible matrices of same size....

You aren't assuming invertibility of ABC. You're just multiplying together 6 matrices, and showing that they give the identity, which is *proving* it is invertible. Just because you wrote "assuming invertibility of ABC" doen't mean you actually did assume it, or that you needed toi.

15. Jul 7, 2008

### b0it0i

Re: Show if A,B,C are invertible matrices of same size....

i think the problem you're having with this proof is understanding what the question is asking.
there are many other proofs with the same structure

the problem is saying show that

(ABC)-1 = C-1B-1A-1

in other words, they want you to show that the inverse of ABC is actually (behaves like) inverse C times inverse B times inverse A

an example would be... Show that

A Source of water = Rain

you must show that rain acts/ behaves like a source of water

apply the properties of "a source of water" to rain

Show that C-1B-1A-1 "behaves" like the (ABC)-1
how do you do that?... you check the property of (ABC)-1

that is... show that C-1B-1A-1 (ABC) = (ABC) C-1B-1A-1 = I

An important fact is to note that "the inverse of a matrix is unique"
since C-1B-1A-1 "behaves" like the inverse of ABC,
and that the inverse of a matrix is unique,
C-1B-1A-1 MUST be (ABC)-1

this is corny, but my past professor constantly called these types of proofs
"walks like a duck, quacks like a duck, MUST be a duck..." proof

16. Jul 8, 2008


# Answer does not make sense using conditional probability

As of now, there are 64.1 million people residing in the UK. 5.4 million of them are thought to be asthmatic. A new test for asthma has recently been introduced and medical trials indicate that

• in patients with asthma, the test correctly returns positive 68% of the time
• in patients that does not suffer from asthma, the test correctly returns 82% of the time.

Assuming a patient undergo medical inspection. What is the probability that the patient has asthma,

1. if the test comes back positive?
2. if the test comes back negative?

For Part 1, I used this method:

Step 1: $$P(A|+)=\frac{P(+|A)P(A)}{P(+)}$$

Step 2: $$P(+)=P(+│A)P(A)+P(+│N)P(N)=(0.68)\left(\frac{54}{641}\right)+(0.18)\left(\frac{587}{641}\right)=\frac{7119}{32050}$$

Step 3: $$P(A│+)=\frac{P(+|A)P(A)}{P(+)}=\frac{0.68\times\frac{54}{641}}{\frac{7119}{32050}}=\frac{204}{791}=25.79\%$$

For Part 2, I used the same method as Part 1 but I got a very small answer. Assuming if I insert all the values in the equations correctly, would it be sensible if my answer for Part 2 is 3.47%?

However, personally, it does not make logical sense if the probability of test returning negative is 3.47% because that would mean almost everyone in the nation would be asthmatic.

Or I could just write it as $$100-25.79 = 74.21%$$ but I'm afraid this isn't the answer given the complexity of the question.

• 3.47% is the right answer here... – Clement C. Apr 29 '15 at 13:33
• You're computing the probability of a false negative not the probability someone doesn't have asthma. False negatives are rare and should be rare intuitively. Otherwise, the test would be a piece of crap. So, you're right. – Jamie Lannister Apr 29 '15 at 13:34
• I don't understand what you mean. If 3.47% is the correct answer, this is quite reasonable. It says: "If the test says you don't have asthma, then it's very likely you really don't have asthma (i.e. the test is good)." The first answer is actually more surprising. It says: "If the test says you have asthma, it's only 25% likely you actually have it!" But that isn't that surprising given the low figure of 68% in the hypotheses. – Frank Apr 29 '15 at 13:34
• @iterence: that would not give the probability $\Pr[A\mid -]$, but instead the quantity $\Pr[A^c\mid +] = 1 - \Pr[A\mid +]$. That is, instead of "the probability to have asthma knowing the test says you don't," you would compute "the probability not to have asthma, knowing the test says you do." – Clement C. Apr 29 '15 at 13:38
• In part 2, "if the test comes back negative", it is assumed we know that the result was negative, i.e., the probability of the test returning negative is $100$%, not $3.47$%. The $3.47$% figure arises from the cases where people do have asthma but it is not detected by the test. – David K Apr 29 '15 at 13:41

One instructive way to do these kinds of $2 \times 2$ problems, since there are only four distinct populations, is to enumerate them and calculate the marginal probabilities by inspection.

$$\begin{array}{|c|c|c|c|} \hline & \text{asthmatic} & \text{not asthmatic} & \text{TOTALS} \\ \hline \text{positive} & 3.672 & 10.566 & 14.238 \\ \hline \text{negative} & 1.728 & 48.134 & 49.862 \\ \hline \text{TOTALS} & 5.400 & 58.700 & 64.100 \\ \hline \end{array}$$

From this, one can read off

$$P(\text{asthmatic} \mid \text{positive}) = 3.672/14.238 \doteq 0.25790$$

and

$$P(\text{asthmatic} \mid \text{negative}) = 1.728/49.862 \doteq 0.34656$$

ETA: Two of the cells—$3.672 = (0.68)(5.400)$ and $48.134 = (0.82)(58.7)$—are filled directly by the given parameters. Everything else is bookkeeping.

The abysmal false positive rate is due to a combination of the low specificity of the test ($82$ percent) and the relatively low prevalence of asthma ($5.4/64.1 \doteq 8.4$ percent), so that although only a minority (sizable, but still a minority) of non-asthmatics test positive, they still dwarf the asthmatics who test positive, because there are so few asthmatics available to test positive in the first place. Even if the test were $100$ percent sensitive, the false positive rate would still have been $10.566/(10.566+5.400) \doteq 66$ percent.


Expected value of sums

Suppose we draw cards out of a deck without replacement. How many cards do we expect to draw out before we get an ace?

-

It depends (slightly) on how one interprets before. There are two interpretations possible: (i) Before means not including the draw that got us the first Ace and (ii) We include in the count the draw that got us the first Ace.

There is no big difference between (i) and (ii): The count (and expectation) in (ii) is just $1$ more than the count, and expectation, in (i), We use interpretation (i). So let $W$ be the number of draws before the first Ace, not including the draw that got us the Ace. We want $E(W)$.

The argument is simple but a bit delicate, so the solution below is given in great detail. Luckily, the actual computation is almost formula-free. We use indicator random variables. Label the $48$ non-Aces $1$ to $48$. Don't bother to label the Aces.

Define random variable $X_i$ by $X_i=1$ if the card with label $i$ was drawn before any Ace, and let $X_i=0$ otherwise. Then $$W=X_1+X_2+\cdots+X_{48}.$$ By the linearity of expectation, which holds even when the random variables are not independent, we have $$E(W)=E(X_1+X_2+\cdots+X_{48})=E(X_1)+E(X_2)+\cdots+E(X_{48}).$$ By symmetry, all the $X_i$ have the same distribution. We find, for example, the probability that $X_1=1$. So we want the probability that card with label $1$ is drawn before any Ace.

Consider the $5$-card collection consisting of the $4$ Aces and the card labelled $1$. All orders of these cards in the deck are equally likely. It follows that the probability that card with label $1$ is in front of the $4$ Aces is $\frac{1}{5}$. Thus $E(X_1)=\frac{1}{5}$.

We conclude that $E(W)=\dfrac{48}{5}$.

If we want to take interpretation (ii), and include the draw that got us the Ace, our expectation is $1+\dfrac{48}{5}=\dfrac{53}{5}$.

-
Now I'm wondering if this answer should be considered essentially the same as mine. – Michael Hardy Nov 27 '12 at 1:54
Hard to decide! Mine (of course it's not mine, it is pretty standard stuff, I have used it in lectures) is very formal. – André Nicolas Nov 27 '12 at 1:59
Has anyone written an exposition of the fact that expectations are often easier to find than probabilities? – Michael Hardy Nov 27 '12 at 2:01
Don't know of one. I sure have stressed it. – André Nicolas Nov 27 '12 at 2:04

Distribute the $52$ cards uniformly between $0$ and $1$, so on average they're at $k/53$ for $k=1,2,3,\ldots,52$. The four aces are on average at $1/5,\ 2/5,\ 3/5,\ 4/5$. So $$0.2 = \frac{k}{53}$$ implies $$k = 10.6.$$ and $(0.8)\cdot53 = 42.4$. So on average the four aces are the $10.6$th, $21.2$th, $31.8$th, and $42.4$th cards.

-

A good way to go about this is as follows. Let $T$ be the number of cards drawn at the time of the first ace. This is a $\mathbb{N}$ valued random variable. Therefore $$E(T) = \sum_{n=0}^\infty P(T > n).$$ You may find this helpful.

-

I think that the answer lies in the Hypergeometric distribution or else k successes in n draws from a finite population of size N containing m successes without replacement (quoting from Wikipedia).

In your case, you want a single success in $p$ draws, and from the distribution you can find an average.

Edit: $$P(X=1) = \frac{\binom{m}{1}\binom{N-m}{n-1}}{\binom{N}{n}}$$ and taking an average should be something like the following $$\bar M = \sum_{i=1}^{N-m}i\frac{\binom{m}{1}\binom{N-m}{i-1}}{\binom{N}{i}}$$ for $N=52, m=4$.

End of edit.

Wikipedia also has some examples with urns, black and white balls, so you can work the answer from those.

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I came across this in The Theory of Gambling and Statistical Logic By Richard A. Epstein so pasting a snippet here. I got this from Google Books (public domain) so hopefully there is no copyright violation here. It's amazing how Andre, Michael and Patrick derived the correct answer differently.

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I did not see this approach in any of the answers, which is one that makes sense to me: you can get the ace in either the 1st, 2nd,..., 49th trial.

Then the expected number of cards dealt is $$1(1/13)+2(12/13)(1/13)+..n(12/13)^{n-1}(1/13)+48(12/13)^{47}(1/13)$$.

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Try this. (edited based on comment below) You will need just 1 pull with probability 4/52 2 with probability (4/51 x 48/52). 48/52 for the probability of not getting and ace in the first and getting it in the second (4/51), and so on. This yields 1 x 4/52 + 2x(4/51 x 48/52) + 3x(4/50 x 48/52 x 44/50) + ... There must be some simplification for the above, but I'ven't looked deeper into, but you get the idea. Here are similar ones

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This assumes there is only one ace in the deck. – Michael Hardy Nov 27 '12 at 1:05

I have always liked this shortcut (proof here), which also holds for continuous random variables defined on $x \geq{0},\,x \in \mathbb{R}$. For some reason it was called "The Darth Vader" rule in my actuarial study manual but I digress. For this problem: $$P(T>n) = \frac{\binom{48}{n}n!}{\binom{52}{n}n!} = \frac{\binom{48}{n}}{\binom{52}{n}}$$ This is easily reasoned as follows:

$P(T>n)$ is the probability that there are no aces up to and including the $n$th draw which means that in these first $n$ draws, out of the $48$ non-aces we count the number of ways to choose $n$ of them and permute them. We divide this by the total number of outcomes for the first $n$ draws which is $\binom{52}{n}n!$ (we must include the aces this time, thus the 52).

Therefore: $$E(T) = \sum_{n=0}^{48}\frac{\binom{48}{n}}{\binom{52}{n}} = \sum_{n=0}^{48}\frac{(52-n)(51-n)(50-n)(49-n)}{52\times51\times50\times49} = 53/5 = 10.6$$

We could sum to $\infty$ like ncmathsadist stated but the values of $P(T>n)$ for $n>48$ are $0$ anyway.

I will admit that I cheated in my final step by using wolframalpha. Wolfram Alpha is a GREAT free resource by the way. I do hope that the logic behind solving the problem is not lost as a result. - Patrick

Sidebar: My initial naive guess was that $E(T) = 13$ since there are on average $12$ cards between each ace. I am trying to figure out an intuitive understanding for why the actual answer is less that $13$ now.

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It is 13 only if there is replacement. As there isn't any replacement the probability goes up so that interval should shrink eventually converging to 10.6. BTW could you elaborate how you simplified the average to 53/5? – broccoli Nov 27 '12 at 7:52
You're right, if there is replacement then $T \sim Geometric(p)$ where $p$ is the probability of finding an ace ($1/13$) and $E(T) = 1/p = 13.$ As for an elaboration I just plugged the sum into wolfram alpha's calculator. Andres solution above is much better as the calculations can be done in your head and his method is applicable to other similar problems. – Patrick Nov 27 '12 at 13:33
If you wonder why $13$ is too big, consider the expected number of trials needed to get all four aces. Would that be $52$? Only if you never get the fourth ace until the last card in the deck. – Michael Hardy Nov 27 '12 at 19:16

Add a fifth ace, uniformly randomly arrange the $53$ cards in a circle, and break the circle into a line at the added ace. By symmetry, the expected number of cards between two of the $5$ aces is $\frac{48}5$, so this is the expected number of cards between the beginning of the line and the first ace.

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# What's the chance that exactly 2 2s will have appeared before the 2nd 8?

A standard deck of 52 cards is randomly shuffled. Bob keeps drawing from the deck until he has drawn two 8s.

What is the probability that, when, he draws the second 8, he has already drawn exactly two 2s?

I think this could be a combinations problem because the order of the drawing two 2s does not matter. However, I am not sure how to account for the fact that he could draw an 8 or a second 8 at any point.

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Hint: one approach is to recognize that all the other cards are just fluff. There are only ${8 \choose 4}=70$ possibilities for the order of 2's and 8's, so list them and count. You can make it less by listing the orders that get to 3 2's or 2 8's and assessing the probability. The first are failures, the second are successes if you have two 2's already. For example, one is that you start with three 2's. This is $\frac 12 \cdot \frac 37 \cdot \frac 26=\frac 1{14}$. Another is 2828. This is $\frac 12 \cdot \frac 47 \cdot \frac 12 \cdot \frac 35=\frac 3{35}$

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Oh! I think I got it from that hint! The only way for this to hold is for the string of combos be 8228[some ordering of 2 2s and 2 8s], 2828[some ordering of 2 2s and 2 8s], or 2288[some ordering of 2 2s and 2 8s]. Hence, the solution is 3 * (4 choose 2) / (8 choose 4), which is approximately 0.257! :D ... right? –  David Faux Sep 9 '12 at 4:03
@DavidFaux: Good thought, but I wouldn't be sure that all the cases are equally probable. 8228 is $\frac 48 \frac 47 \frac 36 \frac 35=\frac 3{35}$ while 2288 is $\frac 48 \frac 37 \frac 46 \frac 35$, which is the same, so I think it works. I would report the result as $\frac 9{35}$, which is exact. –  Ross Millikan Sep 9 '12 at 4:20
Hmm, I think the combinations are equal because non-2 and non-8 cards are all fluff. If we make the 8s non-unique, and 2s non-unique, this basically becomes a problem involving strings of four 8s and four 2s, right? –  David Faux Sep 9 '12 at 4:34
@DavidFaux: That is right. The more I think, I agree with your calculation. I was worried that 22228888 would be less probable than 82282882, but now I don't think so. It is easy to think things are equally probable when they are not. –  Ross Millikan Sep 9 '12 at 4:39

As was observed by Ross Millikan, there are only $8$ relevant cards. The first $4$ of these can appear in the dealing in $(8)(7)(6)(5)$ orders, all equally likely.

We count the number of "favourable" orders. These are given by any one of three patterns, which we call $2288$, $2828$, and $8228$.

The pattern $2288$ can occur in $(4)(3)(4)(3)$ ways, as can the other two patterns. Thus the required probability is $$\frac{(3)(4)(3)(4)(3)}{(8)(7)(6)(5)}.$$ If we want to simplify a bit, we get $\dfrac{9}{35}$.

Remark: The problem seems to have been solved correctly in the comments both by you and by Ross Millikan. Then there appears to have been a change of mind.

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# Electric field of bent non conducting rod

## Homework Statement

You are given a non conducting rod carrying uniformly distributed charge, -Q, that has been bent into a 120° circular arc of radius, R. The axis of symmetry of the arc lies along the x-axis and the origin is at the center of curvature of the arc.
(a) in terms of Q and R, what is the linear charge density, λ?
(b) in terms of Q and R, what is the magnitude and direction the the resulting electric field at the origin?
(c)If the arc is replaced by a point particle carrying charge, -Q, at x=R, by what factor is the resulting electric field at the origin multiplied?

λ=Q/L
E=KQ/R2
dq=λds
ds=Rdθ

## The Attempt at a Solution

(a) λ=Q/L, λ=Q/(⅓2piR) = 3Q/2piR

(b) Cant seem to figure out how to write an integral with bounds. The bounds on my integral below will be from θ=0° → θ=60°
E=2∫dEx=2∫kλdθcosθ/R = 2kλ/R ∫cosθdθ (Like previously stated this integral is being evaluated from θ=0° to θ=60°)
E=(2kλ/R)[sin60°-sin0°] = (kλ√3)/R, it wants in terms of R and Q so now plug in our above statement for λ
E=(3kQ√3)/2piR2

(c) Im a little unsure about what this is asking, from my understanding:
to obtain the new electric field from a point particle which is simply kQ/R2 we must multiply the previous electric field from bent rod by a factor of 2pi/3√3. Not sure if this is correct or if I'm thinking about the question in the wrong way.
Any help would be much appreciated
Thanks!

Related Introductory Physics Homework Help News on Phys.org
kuruman
Homework Helper
Gold Member
Consider subdividing the 120o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.

Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?

Consider subdividing the 120o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.

Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?
Okay, Ill work on this and post my results! Thanks!

Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?

Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?
Yea just part c, wasn't sure what it was asking but @kuruman helped me out with that one. Just have a test in a few days and want to make sure I'm doing these problems correctly. Always love reassurance! LOL

You saw the symmetry and ignored the y component. You also used symmetry to cut the integral in half. You integrated in theta and didn’t forget the extra R.

I don’t think you have much to worry about, but I’ll wish you good luck anyway.


# Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$

Compute the limit $$\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$$

How can this be done? The best I could do was rewrite the limit as $$\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n}$$

Following that log suggestion in the comments below: \begin{align} &\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\ &= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\ &= \int_1^2 \ln x \, dx \\ &= (x \ln x - x)\vert_1^2 \\ &= (2 \ln 2 - 2)-(1 \ln 1-1) \\ &= (\ln 4-2)-(0-1) \\ &= \ln 4-1 \\ &= \ln 4 - \ln e \\ &= \ln \left( \frac 4e \right) \end{align}

but I read from somewhere that the answer should be $\frac 4e$.

• Your integral is the log of the answer – robjohn May 14 '18 at 18:15
• @CarlMummert From the accepted answer of this meta question, we can see that this is a well-written question (except the use of $\frac1n$ as a power), and "additional indication of source would not make a difference".You may re-read AmateurMathGuy's reply to your comment to know why they are not required. – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
• You voted to close because "Being homework is a sufficient reason to vote to close a question", but HW questions are allowed. – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
• @CarlMummert I do not see how my question "has no real context, and does not fit the quality standards that many users expect." I put in the work and effort as this Math SE site suggests. – GarlicBread May 15 '18 at 3:58
• Agreed. This is supported by the Close Queue Review result. All reviewers left this question open. – GNUSupporter 8964民主女神 地下教會 May 15 '18 at 9:00

Hint: let $a_n=(n+1)\cdots(2n)/n^n$, and use the fact that $a_{n+1}/a_n\to L$ implies $\sqrt[n]{a_n}\to L$ (i.e., the ratio test implies the root test).

If you know Sterling's Approximation: $$n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n},$$ then you could approach it as follows:

$$\lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{\left(\frac{2n}{e}\right)^{2n} \sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{2^{2n} \left(\frac{n}{e}\right)^n \sqrt{2}} = \lim_{n \to \infty} \frac{1}{n} \cdot 2^2 \cdot \frac{n}{e} \cdot 2^{\frac{1}{2n}} = \frac{4}{e}.$$

Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.)

• I'm not sure how to rewrite the limit of the Riemann sum as its corresponding definite integral, although I think I managed to arrive at the Riemann sum in my work. – GarlicBread May 14 '18 at 17:41

If you accept using the following limit that has been asked and proved (for example here) many times here on MSE

• $\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,

then your limit is easily derived:

$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$

Note that $\displaystyle\frac{\sqrt[n]{n+1}}{n}\neq\left(\frac{n+1}{n}\right)^{1/n}$

Also, this is just for clarification since some nice suggestions have already been given but also because the question mark remains over your last equality

With this is mind and going from your first step,

$$S=\lim_{n\to\infty}\frac{\sqrt[n]{(n+1)(n+2)...(2n)}}{n}=\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$

$$\operatorname{ln}(S)=\operatorname{ln}\left(\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\operatorname{ln}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}\right)+\operatorname{ln}\left((n+2)^{1/n}\right)+...+\operatorname{ln}\left((2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\operatorname{ln}\left((n+p)^{1/n}\right)\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\frac{1}{n}\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}\left(1+\frac{p}{n}\right)+\operatorname{ln}(n)\right)-\operatorname{ln}(n)\right)$$

This $+\operatorname{ln}(n)$ is independent of $p$ and summed $n$ times, then divided by $n$, so $\operatorname{ln}(n)-\operatorname{ln}(n)=0$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)$$

This is what Eric meant and the argument varies from $1$ to $2$

It becomes a little clearer if we let $m = 1/n$

$$\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)=\lim_{m\to\ 0}\left(\left(\sum_{p=1}^{1/m}\operatorname{ln}(1+pm)\right)m\right)=\int_1^2\operatorname{ln}(x)dx=(x\operatorname{ln}(x)-x)\Big\rvert_1^2$$

$$\therefore \operatorname{ln}(S)=2(\operatorname{ln}(2)-1)+1=\operatorname{ln}(4)-1=\operatorname{ln}\left(\frac{4}{e}\right)$$

Finally

$$S=\frac{4}{e}$$