Chapter 13.2, Problem 24HP

To determine

To prove: that relation of permutation and combination $r!C{}_r=P{}_r$ is true and explain why it is true.

Answer to Problem 24HP

The equation $r!C{}_r=P{}_r$ is proved.

Explanation of Solution

Given:

  $r!C{}_r=P{}_r$.

Concept used:

According to the combination formula:

  $C{}_r=\frac{n!}{r!\left(n-r\right)!}.$

According to the permutation formula:

  $P{}_r=\frac{n!}{\left(n-r\right)!}.$

Calculation:

  $C{}_r$ and  $P{}_r$  .

Differ by the factor $r!$  because there are always  $r!$  ways to order the groups that are selected. Therefore, there are  $r!$  permutations of each combination. For example, there are  $6$  combinations of  $4$  letters taken  $2$  at a time 

  $\left(AB, AC, AD, BC, BD, CD\right)$ in this case  $r=2$ because that is how many objects are taken at a time. There are also  $2!$  permutations of each combination.

  $\left(AB/BA, AC/CA, AD/DA, BC/CB, BD/DB, andCD/DC\right)$ .

  $\begin{array}{l}r!C{}_r=r!\frac{n!}{\left(n-r\right)!r!}\\ \Rightarrow \frac{n!r!}{\left(n-r\right)!r!}.\\ \Rightarrow \frac{n!}{\left(n-r\right)!}=P{}_r.\end{array}$

Hence, $r!C{}_r=P{}_r$ .