You have n boys sitting on a bench. 2 of them can not sit at the end. A different 2 must sit together How many ways are there of achieving this? I was quite pleased with coming up with 2(n-2)(n-3)(n-3)! in a few minutes, until I found out it was wrong. This was more-or-less my [flawed] reasoning, which I hope makes sense: Let's consider the 2 boys who can't sit in the middle. There are n-2 positions the first one can go in. The second can go in n-3 positions. So far that's (n-2)(n-3) ways of arranging those 2. We now have n-2 boys left to place. But two of them have to sit together - so let's call those one unit - so we consider how to place n-3 objects. So that's (n-3)! ways (x 2 of course to allow for the congealed boys to swap places). So putting it altogether, you get 2(n-2)(n-3)(n-3)! Which is wrong. The problem is, after I'd thought about it, there aren't (n-3)! x 2 ways of arranging the final n-2 boys. There would be if there were n-2 empty places next to each other, but that of course is not the case. And this is the whole problem - depending on WHERE the first 2 boys go, affects how many ways you can position the remaining n-2 boys. So you can't say for every x positions the first 2 go in, there are y others for the remaining n-2, therefore the answer is XY. I spent hours on this, and still can't get it. Yet I can't have been that far out in the first place,as the correct answer is: 2(n-3)(n-4)(n-3)! which at least looks like my first attempt. Please can you explain how to get this. Thanks very much.