[Math] 4 men and 4 women are to be seated in a row. A man and a woman are dating and must be seated together. How many ways can this be done

combinatorics

I was helping someone with their math homework and am having a lot of trouble with this question. I was wondering if anyone can please help me.

$\mathbf {Question:}$

4 men and 4 women are to be seated in a row. A man and a woman are dating and must be seated together. How many ways can this be done?

$\mathbf {Attempt at Solution}$

At first, I thought that there are 4*4 = 16 ways to make a couple. We count the couple as one object so there are now 7 objects to rearrange. So the answer seems to be 7!*16. But then this exceeds the number of ways to arrange the 4 men and 4 women (with no restrictions) which is 8!

The problem, I realize, is that we are double counting some of the couples. When we pick a couple and move them around (and count), then pick another couple and move them around (and count), it will happen that a couple we have counted before will be counted again.

Can someone please help me with this?

Thank you!

(PS: Sorry for my bad Latex-ing, it's been some time.)

Best Answer

The couple must be placed togther, and there are 7 possible adjacent positions (1 and 2, 2 and 3, ..., 7 and 8), and once you select that, 2 ways to place the couple. The other 6 people can be placed arbitrarily in the remaining 6 spots, and going through, the first can be placed 6 ways, the next 5, the next 4, etc. Therefore, there are $$(7)(2)(6!)=(2)(7!)=10080$$ possible ways to place everybody.

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