m men and n women are to be seated in a row s | vedclass

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(A) First,arrange $m$ men in a row in $m!$ ways.Since $n < m$ and no two women can sit together,we consider the gaps between the men.In any one of the

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

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(A) First,arrange $m$ men in a row in $m!$ ways.Since $n In any one of the $m!$ arrangements,there are $(m + 1)$ available spaces (including the ends) where $n$ women can be placed.The number of ways to arrange $n$ women in these $(m + 1)$ spaces is given by $^{m+1}P_n$.Therefore,by the fundamental principle of counting,the total number of arrangements is $m! 	imes {}^{m+1}P_n$.$= m! 	imes \frac{(m + 1)!}{(m + 1 - n)!} = \frac{m! (m + 1)!}{(m - n + 1)!}$.

$m$ 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

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