Question

$m$ 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?

• A. $\frac{m!\left(m+1\right)!}{\left(m-n+1\right)!}$
• B. $\frac{m!\left(m-1\right)!}{\left(m-n+1\right)!}$
• C. $\frac{\left(m-1\right)\left(m+1\right)!}{\left(m-n+1\right)!}$
• D. None of these

Answer 1

The correct option is A

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

Explanation for the correct answer:

Finding the seating arrangement:

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

$m$ men can be arranged to sit in $m!$ ways

Since no two women can sit together then, there are only $m+1$ places available for women to sit.

so, $n$ women can be arranged to sit in $m+1$ places in ${}^{m+1}P_n$ ways

So, the Total Number of ways in which $m$ men and $n$ women are to be seated in a row so that no two women sit together are

$=m!\times {}^{m+1}P_n$

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

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

Hence, option (A) is the correct answer.