m men and n women are to be seated in a row so that no two women sit together.

Question:

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

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

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

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

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