# 11 books, all with different titles, are to be arranged on a shelf. 4 are history books and one of them needs to be in the middle of the arrangement. 2 are math books and they need to be on the ends. How many different ways can they be arranged?

$3 , 386 , 880$

#### Explanation:

I'm going to write this assuming that the books are distinguishable (and I think this is a fair assumption given the wording of the question indicating that the books are all different).

Because we're talking about the way to arrange books that are distinguishable, I'm going to be using both factorials and the permutation formula, the general form of which is:

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

Let's first look at the books that we know are "locked" in place - there is a history book in the middle of the shelf. How many different choices do we have for that book? ${P}_{4 , 1} = 4$

There are also two math books that we need to have locked in - one at either end. How many different ways can we set down these two math books? P_(7,2)=(7!)/((2!)(5!))=(7xx6xx5!)/(2xx5!)=21

With three books now out of the way, we can address the remaining eight books. We can place the remaining eight any way we want. We can express this as P_(8,8)=8! =40,320

We can now multiply all these together:

$4 \times 21 \times 40320 = 3 , 386 , 880$