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A First Course in Probability

Sheldon M. Ross

$Math Studyset Vaia Explanations$ Math

9th Edition · 432 pages

ISBN: 9780321794772

Chapter 1: Q. 1.7 (page 17)

Give an analytic proof of Equation (4.1).

Short Answer

Expert verified

It is proved that $(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ n - r \end{matrix})$ .

Step by step solution

01

Step 1. Given information.

We have to prove the combinatorial explanation of the identity, that is $(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ n - r \end{matrix})$ .

02

Step 2. Prove the combinatorial explanation of the identity.

The combinatorial explanation of the identity is $(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ n - r \end{matrix})$ .

Taking L.H.S, we have $(\begin{matrix} n \\ r \end{matrix})$ .

Since role="math" localid="1647855220565" $(\begin{matrix} x \\ r \end{matrix}) = \frac{x!}{r! (x - r)!}$ , so

$(\begin{matrix} n \\ r \end{matrix}) = \frac{n!}{r! (n - r)!}$

Therefore, L.H.S = role="math" localid="1647855382907" $\frac{n!}{r! (n - r)!}$

Taking R.H.S, we have $(\begin{matrix} n \\ n - r \end{matrix})$

Since $(\begin{matrix} x \\ r \end{matrix}) = \frac{x!}{r! (x - r)!}$ , so

$(\begin{matrix} n \\ n - r \end{matrix}) = \frac{n!}{(n - r)! (n - n + r)!} = \frac{n!}{r! (n - r)!}$

Therefore, R.H.S $= \frac{n!}{r! (n - r)!}$

As L.H.S = R.H.S, hence it is proved that $(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ n - r \end{matrix})$ .

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Most popular questions from this chapter

Determine the number of vectors $(x_{1}, . . ., x_{n}),$ such that each $x_{i}$ is either $0$ or $1$ and $\sum_{i = 1}^{n} x_{i} \geq k$ .

In how many ways can $r$ objects be selected from a set of $n$ objects if the order of selection is considered relevant?

A committee of $7$ , consisting of $2$ Republicans, $2$ Democrats, and $3$ Independents, is to be chosen from a group of $5$ Republicans, $6$ Democrats, and $4$ Independents. How many committees are possible?

Delegates from $10$ countries, including Russia, France, England, and the United States, are to be seated in a row. How many different seating arrangements are possible if the French and English delegates are to be seated next to each other and the Russian and U.S. delegates are not to be next to each other?

From a group of $8$ women and $6$ men, a committee consisting of $3$ men and $3$ women is to be formed. How many

different committees are possible if

(a) $2$ of the men refuse to serve together?

(b) $2$ of the women refuse to serve together?

(c) $1$ man and $1$ woman refuse to serve together?

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