Question

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

• A. $\frac{(m+1)!m!}{(m+n+1)!}$
• B. $\frac{(m+1)!m!}{(m+n-1)!}$
• C. $\frac{(m-1)!m!}{(m-n+1)!}$
• D. $\frac{(m+1)!m!}{(m-n+1)!}$

Answer 1

The correct option is D $\frac{(m+1)!m!}{(m-n+1)!}$

$m$ men can be seated in $m!$ ways, creating $(m+1)$ for ladies to sit $n$.
Ladies out of $(m+1)$ places (as n < m) can be seated in ${}^{m+1}{P}_m$ ways.
Therefore, total number of ways is
$=m!{\times }^{m+1}{P}_n=m!\times \frac{(m+1)!}{(m+1-n)!}=\frac{(m+1)!m!}{(m-n+1)!}$