how many ways can 18 different books be given to 4 people so that 1 person has 6 books, 1 has 2, and 2 have 5?

Answer 1

The first person has 18 books and chooses 6. This leaves 12 books.
The second person has 12 books and chooses 2. This leaves 10 books.
The third person has 10 books and chooses 5.
The last person gets the last 5 books.

The long formula would be:
(18 choose 6) * (12 choose 2) * (10 choose 5) * (5 choose 5)

n choose k = n! / k! (n-k!)

18! / (6! * 12!) x 12! / (2! * 10!) x 10! / (5! * 5!) x 5! / (5! * 0!)

If you write out the formula it reduces to:
18!
--------------
6! 2! 5! 5!

= 308,756,448 ways

Answer 2

This will call for combinations, since the order that one receives is not really relevent.
C(18,6)*C(12,2)*C(10,5)*C(5,5)
Notcie we started with 18 books from which we were selecting 6 for person #1,C(18,6)
After which we had only 12 books from which to select 2 for person #2, C(12,2)
After those two we were left with 10 books to select 5 for person #3, C(10,5)
Finally the last person gets whatever is left, really only one selection which is what C(5,5) equals.
The reason these selections are multiplied is because, for any one selection one will have to consider all the possible selections for the subsequent people as another way to distribute the books.
I will leave it to you to evaluate.