Chapter 1: Q. 1.17 (page 19)

Present a combinatorial explanation of why
$(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ r, n - r \end{matrix})$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

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

Step 2. Give explanation for nr=nr, n-r

$(\begin{matrix} n \\ r \end{matrix})$ represents the no. of ways in which $r$ objects can be selected from $n$ objects.

When this is done, $r$ objects that are selected are grouped separately and $(n - r)$ objects that are not selected are grouped separately.

Hence, this is equivalent to separating $n$ objects into two groups, one having $r$ objects and another having $(n - r)$ objects.

This can be done in $\frac{n!}{r! (n - r)!} ways = (\begin{matrix} n \\ r, n - r \end{matrix})$

Therefore, $(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ r, n - r \end{matrix})$ .

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Present a combinatorial explanation of why
$(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ r, n - r \end{matrix})$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Give explanation for nr=nr, n-r

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Step by step solution

Step 1. Given information.

Step 2. Give explanation for nr=nr, n-r

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Most popular questions from this chapter

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Present a combinatorial explanation of why
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