Chapter 9.6, Problem 41E

To determine

Find the different batting orders which are possible.

Answer to Problem 41E

The different batting orders which are possible ${}_{15}{P}_9=\frac{15!}{(15-9)!}=1816214400$

Explanation of Solution

Given:

A baseball coach is creating a mine-player batting order by selecting from a team of 15 players

Since the order of batting matters here, this must be a permutation.

There are ${}_{15}{P}_9$ ways of arranging 15 team players for 9 player batting order.

Otherwise this can also be done using fundamental counting principle. For the first batting order there are 15 players available, for 2nd position 14 players are available, for 3rd .....for 9th there are 7 players available.

So batting order could be done in

  $\begin{array}{l}15\times 14\times \cdots \times 7=1816214400\hfill \\ {}_{15}{P}_9=\frac{15!}{(15-9)!}=1816214400\hfill \end{array}$