Question

$m$ 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

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

Answer 1

Answer: (A)

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

$\to m$ men can be arranged in $m$ seats $=m!$ ways.

$-$ After $m$ men be seated, we mat get one seat in the beginning and one seat in the end and $(m-1)$ seats bet $n$ each pair of men for the women .

$-$ Thus the no. of ways which $n$ women can be seated in $(m-1+1+1)=(m+2)$ seats $=m+1p_n$

$-$ total numbers of arrangement $=\left(m\right)!\times m+2p_n$

$=\frac{(m!)(m+1)!}{(m+2-n)!}$

$=\frac{(m!)(m+1)!}{(m.n+D)!}$