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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 show that the number of ways in which they can be seated as m ! ( m + 1 ) ! ( m − n + 1 ) ! - Mathematics

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+1Pn = $\frac{\left( m + 1 \right)!}{\left( m + 1 - n \right)!}$

∴ 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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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 16 Permutations
Exercise 16.4 | Q 10 | Page 37
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