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

Answer 1

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