*m* men and *n* women are to be seated in a row so that no two women sit together. if *m* > *n* then show that the number of ways in which they can be seated as\[\frac{m! (m + 1)!}{(m - n + 1) !}\]

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#### Solution

'*m*' men can be seated in a row in *m*! ways.

'*m*' men will generate (*m*+1) gaps that are to be filled by '*n*' women = Number of arrangements of (*m*+1) gaps, taken '*n*' at a time = ^{m+}^{1}*P _{n}* = \[\frac{\left( m + 1 \right)!}{\left( m + 1 - n \right)!}\]

∴ By fundamental principle of counting, total number of ways in which they can be arranged =\[\frac{m!\left( m + 1 \right)!}{\left( m - n + 1 \right)!}\]

Concept: Factorial N (N!) Permutations and Combinations

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