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$m$ 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
$\frac{{m\;!\;(m + 1)\;!}}{{(m - n + 1)\;!}}$
$\frac{{m\;!\;(m - 1)\;!}}{{(m - n + 1)\;!}}$
$\frac{{(m - 1)\;!\;(m + 1)\;!}}{{(m - n + 1)\;!}}$
None of these
Solution
(a) 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 $^{m + 1}{P_n}$ ways.
$\therefore \,$By the fundamental theorem, the required number of arrangements of m men and n women $(n < m)$
= $m\,!{.^{m + 1}}{P_n} = \frac{{m\,!.(m + 1)!}}{{\{ (m + 1) – n\} \,!}}\, = \frac{{m!(m + 1)!}}{{(m – n + 1)\,!}}$.
