Better Explanation for an already posted question

Question

Can anyone explain why in this question the answer is 5! * 2! * 10P3? I understand the 5! and 2! but for 10P3 the first thing I thought of was 3!

Thanks.

Answer 1

${^{10}\mathrm P_3}$ counts the permutations of a selection of $3$ items from a set of $10$ distinct items.

$3!$ just counts the permutations of $3$ distinct items.   You also have to count ways to select the items.

$${^{10}\mathrm P_3} = {^{10}\mathrm C_3}\cdot 3! = \frac{10!}{7!}$$

In any case, it is not exactly used to answer the posed question.

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There are five distinct computer science books, three distinct mathematics books, and two distinct art books. In how many ways can these books be arranged on a shelf if one of the art books is on the left of all computer science book, and the other art book is on the right of all the computer science books?

You want arrangements like _A_C_C_C_C_C_A_, where the 3 math books can be placed in any of these 8 spaces before, between, and after the art and computer science books. So we count ways to arrange the computer science books in their fixed places, arrange the art books in their places (one either side of the CS books), select 3 from among 8 places for three math books and arrange them with in those places.

You also have to account for cases where two or three math books are placed into the one spot (while preserving their already selected order).

$$5!\;2!\;3!\;\bigg({^8\mathrm C_3}+2\;{^8\mathrm C_2}+{^8\mathrm C_1}\bigg)$$

Which is numerically equal to $5!2!\;{^8\mathrm P_3}+\ldots$ but does not quite reflect the logic used to obtain the count.