Question

It is required to seat $5$ men and $4$ women in a row so that the women occupy the even places. How many such arrangements are possible$?$

Answer 1

Step $1$: finding number of ways of arranging $4$ women in even places

There are $9$ places to be filled in which $4$ are even and $5$ are odd.

Total number of women $=4$

$\therefore$ number of ways arranging $4$ women in $4$ even places

$={}^4{P}_4$

$=\frac{4!}{(4-4)!}=\frac{4!}{0!}$

$=\frac{4\times 3\times 2\times 1}{1}$
$=24$

Step $2$: Finding number of ways arranging $5$ men.

Total number of men $=5$

Total number of ways of arranging $5$ men in remaining $5$ places= $={}^5{P}_5$

$=\frac{5!}{0!}$

$=5\times 4\times 3\times 2\times 1=120$

So, total number of arrangements $=24\times 120=2880$

Hence, total number of arrangements $=2880$