First slide

Permutations

Question

$m$ men and $w$ women are to be seated in a row so that no two women sit together. If $m > w,$ then the number of ways in which they can be seated is:

Moderate

A

$\frac{m! (m + 1)!}{(m - w + 1)!}$
B

$^{m} C_{m - w} (m - w)!$
C

$^{m + w} C_{m} (m - w)!$
D

none of these

Solution

We first arrange the m men. This can be done in $m!$ ways. After m men have taken their seats, the women must choose w seats out of $(m + 1)$ seats marked with X below.
$X M X M X M X \dots X M X$
$1st 2nd 3rd m th (m + 1) th$
They can choose $w$ seats in $^{m + 1} C_{w}$ ways and arrange w women in $w!$ ways.
Thus, the required number of arrangements is
$m! (^{m + 1} C_{w}) (w!) = \frac{m! (m + 1)! w!}{w! (m + 1 - w)!} = \frac{m! (m + 1)!}{(m + 1 - w)!}$

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