Question

if m men and n women are to be seated in a row so that no two women sit together. If m > n, then , the number of ways in which they can be seated is $\frac{m!(m+1)!}{(m-n+1)!}$

• A. True
• B. False

Answer 1

The correct option is A True

$m$ men can be arranged in $m!$ ways.

As no two women are to be together, we have $m+1$ places for women.

Therefore the women can be seated in ${}^{m+1}{P}_n$.

$\therefore$ total no. of ways of seating men and women $=m!\left({}^{m+1}{P}_n\right)$

As we know that,

${}^n{P}_r=\frac{n!}{(n-r)!}$

$\therefore$ total no. of ways of seating men and women $=m!\left(\frac{(m+1)!}{(m+1-n)!}\right)=\frac{m!(m+1)!}{(m-n+1)!}$

Hence the correct answer is $\left(A\right)\text{True}$.