Question

A group contains n men and n women. How many ways are there to arrange these people in a row if the men and women alternate?

Short answer

There are$2{\left(\mathrm{n}!\right)}^2$ groups the n men and n women.

Step-by-step solution

Given data

There are n men and n women.

Concept of Factorial

Factorial is the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point.

Thus, factorial seven is written as$\mathbf{1}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{6}\mathbf{\times }\mathbf{7}$.

Calculation for number of ways to arrange people

The men can arrange n!ways and similarly the women can arrange ways so . The men and women arrange alternative so two possibilities either start with men or start with women $2\times \mathrm{n}!\times \mathrm{n}!\text{which is}2\times \mathrm{n}!\times \mathrm{n}!=2(\mathrm{n}!{)}^2\text{.}$

Thus, there are ${\left(2n!\right)}^2$groups the n men and n women.