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 –

• A. $\frac{m!(m + 1)!}{(m - w + 1)!}$
• B. ^mC_m–w (m –w)!
• C. ^m+wC_m(m –w)!
• D. None of these

Answer 1

Answer: (A)

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

Answer 2

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……. X M X

1^st 2^nd 3^rd m^th

They can choose w seats in ^m+1C_w ways and take their seats in w! ways.

Thus, the required number of arrangements is

m!(^m+1C_w)(w!)= $\frac{m!(m + 1)!w!}{w!(m + 1 - w)!}$= $\frac{m!(m + 1)!}{(m + 1 - w)!}$