Q. 1.7 Give an analytic proof of Equati... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 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

Step 1. Given information.

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

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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91影视

Give an analytic proof of Equation (4.1).

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Prove the combinatorial explanation of the identity.

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