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Question: m men and n women are to be seated in a row, so that no two women sit together. If \(m > n\), then t...

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

A

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

B

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

C

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

D

None of these

Answer

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

Explanation

Solution

First arrange m men, in a row in m ! ways. Since n < m and no two women can sit together, in any one of the m ! arrangement , there are (m + 1) places in which n women can be arranged in m+1Pnm + 1P_{n} ways.

∴\therefore By the fundamental theorem, the required number of arrangement = m ! m+1Pn=m!(m+1)!(m−n+1)!m + 1P_{n} = \frac{m!(m + 1)!}{(m - n + 1)!}.`