# How many ways can $5$ men and $5$ women be arranged so that you cant have two men and two women stand next to each other?

My reasoning is, there are $5!$ ways to order $5$ men. Then there are 6 gaps between the $5$ men to place the $5$ women, this can be done in $\binom{6}{5} = 6$ different ways. Hence a total of $5!*6 = 720$.

The answer is $2*5!*5! = 28800$.

Could someone please explain what I am doing wrong, I'm stuck.

Could someone please explain what I am doing wrong?

The following method:

• There are $6$ gaps between the $5$ men to place the $5$ women
• This can be done in $\binom65=6$ different ways

Is wrong for two different reasons:

• There are indeed $6$ gaps between the $5$ men, but you cannot choose ANY $5$ of them, because although the women will not be next to each other, some of the men might be
• Once you choose the gaps, you still need to arrange the women in any possible order, so you need to multiply the result of the previous step by the number of ways to do it

To put it simple:

• There are $2$ ways to choose the gaps - either the $5$ leftmost gaps or the $5$ rightmost gaps
• There are $5!$ ways to arrange the women in any possible order within the $5$ chosen gaps

Hence in your answer, you should replace the "$6$" with "$2\cdot5!$".

You have $5!$ possible orders for the men alone, and as much for the women alone.

When you take one order for each sub-group, you then have two possible full orders : either start with the first man, then the first woman, then the second man... or do the opposite, start with the first woman, then the first man...

This gives the result.

You went wrong when you looked at the gaps. The men are women are to be standing in line. And once you have an order for the men, one has to be chosen for the women too

Notice that there is $5!$ ways to order the men. Moreover, there are $6$ gaps and only five women thus we can choose to not utilize a gap. However, we don't have the freedom to choose which gap not to use because if we don't use a gap in the middle two men touch, thus we can only not use gaps on the end of which there are two. Thus we have $2$ different ways of choosing which gap to not use. Upon choosing the gaps we have $5!$ ways to distribute the women among those gaps. Thus in total $2\cdot 5!\cdot 5!$ ways.

The above answers are a direct way of seeing the question. But here's another way to think of it. Group 1 man and 1 woman together, how many ways can we can we do this $5!*5!$. But we can lead with either man or woman first. Giving us two ways.

Thus in total we have $5!*5!*2=28800$