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?

Relevant definitions to solve this exercise:

“THE PRODUCT RULE Suppose that a procedure can be broken down into a sequence of two tasks. If there are n1 ways to do the first task and for each of these ways of doing the first task, there are n2 ways to do the second task, then there are n1n2 ways to do the procedure.”

“THE SUM RULE If a task can be done either in one of n1 ways or in one of n2 ways, where none of the set of n1 ways is the same as any of the set of n2 ways, then there are n1 +n2 ways to do the task.”

The definitions above are from the textbook Discrete Mathematics and its Applications by Rosen.

Let’s assume that we can start always with a woman.

We can choose the first woman in n ways.

After that, we can choose the second place, which is the first man, in n ways.

Then, the third place, which is woman, we can choose it in n-1 ways.

Similarly, the fourth place, which is a man, we can choose it in n-1 ways.

It follows that we can apply the multiplication rule.

nxnx(n-1)x(n-1)x(n-2)x(n-2)x…

Arranging in a convenient way, we get

= n(n-1)(n-2)…x 1 x n(n-1)(n-2)…x 1

= n! x n!

= n!²

So, n!² is the number of ways we alternate n man and n women starting with a woman. So, if we start with a man, we can order them in the same number of ways n!².

Because we can either start with a man or a woman, it follows that in this case we must apply the sum rule.

So, the final answer is:

n!² + n!² = 2 n!²

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