I have 4 different mathematics textbooks and 3 different psychology textbooks. In how many ways can I place the 7 textbooks on a bookshelf, in a row, if there must be a psychology textbook exactly in the middle, and there must be a mathematics textbook at each end?
First we fix the psychology book in the middle. There are 3 choices we can make for this book.
For the remaining 6 slots (3 on either side of the psychology book), we can put any of the 7 textbooks in the first slot.
Once we put a book in the first slot, we have 6 choices left for the second, 5 choices left for the third, and so on.
So, following the fundamental counting principle, we might be tempted to say there are 7⋅6⋅5⋅4⋅3⋅2=(7−6)!7!=5040 ways of arranging the books.
However, this overcounts all the arrangements where we just move all the math books to one side and all the psychology books to the other side.
There are 2! ways of doing this (since we can switch the two sides), so we have to divide our previous answer by 2.
Therefore, there are 7!/(7- 6)! * 1/2! = 2520 ways of arranging the textbooks.