A Chef has $15$ cookbooks. In how many ways can he choose $8$ books and line them up on a shelf above his kitchen counter? 
A Chef has $15$ cookbooks. In how many ways can he choose $8$ books and line them up on a shelf above his kitchen counter?

I solved it the following way:
Ways of choosing $8$ books out of $15$ books:
$15C8= \frac{15!}{(15-8)!(8!)} \implies 15C8=6435$ ways
Then I calculated for the ways to arrange those $8$ books in a line:
$8!=40320$ ways
Lastly, to get the total number of ways:
$6435 \cdot 40320 = 259459200$ ways
But I am not sure if I did this the correct way.
 A: The way the question is put, "choose books and line them up" your answer is the expected way,  $^{15}C_8 \times 8!$
but you should brush up your fundamentals
to realize that $^{15}C_8\times 8!$ is the same as $^{15}P_8$
which, of course can be written as $(15*14*13*12*11*10*9*8)$
A: My Solution: (it may be helpful to you)


*

*Case 1: If the order of different cookbooks is cared (i.e, $ABC\ne BCA, \cdots$), then:

(1) You can choose 1 cookbook from the remained 15 cookbooks and put it on the shelf.
(2) You can choose 1 cookbook from the remained 14 cookbooks and put it on the shelf.
...
(8) You can choose 1 cookbook from the remained 8 cookbooks and put it on the shelf.
So the way number is $P(15,8)=15\times 14\times \cdots \times 8=259459200$.


*

*Case 2: If the order of different cookbooks is not cared (i.e, $ABC=ACB=BAC=BCA=CAB=CBA, \cdots$), then the total way number should be divided by the permutation number of $8$ cookbooks.

So the way number is $C(15,8)=(15\times 14\times \cdots \times 8)/(8\times 7\times \cdots \times 1)=6435$.
A: There are 15 books we can choose to put first on the bookshelf. There are then 14 books we can choose to put second- so there are 15(14) different ways to choose the first two books.  There are then 13 ways to choose the third book, 12 ways to choose the fourth book, 11 ways to choose the fifth book, 10 ways to choose the sixth book,9 ways to choose the seventh book, and 8 ways to choose the eighth book.  So there are 15(14)(13)(12)(11)(9)(8).  That can be written Z as 15!/7!
