SOLUTION: How many ways can seven books be arranged on a shelf if one of the books is an algebra text and must be in the center?

Question 765892: How many ways can seven books be arranged on a shelf if one of the books is an algebra text and must be in the center?
Answer by MaartenRU(13) (Show Source): You can put this solution on YOUR website!
This is a complicated way to ask a simple question. Let's have a look, part by part.

How many choices do you have for the first (leftmost) book? Well, there's seven books, but we have to put the algebra book in the middle, so we can't choose that one. So there's six choices left.

Now, for the second book, we still can't choose the algebra book, but now we also can't choose the first book we chose - that one's already on the shelf. So we can choose five.

For the same reason the third book has four possible choices.

The fourth book is the algebra book, so we don't really have a choice here.

We'll go on like this, so for the fifth book we have three choices, for the sixth we have two, and for the last one we have no choice, because there's only one book left.

Multiplying all these choices together gives us:

6*5*4*1*3*2*1=720. You might notice that this is the exact same number as when we would arrange 6 books on a shelf. That's because we're doing exactly that! We're arranging our six books on a shelf, and sticking an algebra book in the middle. Since the algebra book has a fixed position, we don't have to include it in our calculation.
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