Question

An English teacher needs to pick 1111 books to put on his reading list for the next school year. He has narrowed down his choices to 1212 novels, 88 plays, and 1212 nonfiction books. If he wants to include 55 novels, 44 plays, and 22 nonfiction books, how many ways can he choose the books to put on the list?

Answer 1

There are 3,659,040 ways he can choose the books to put on the list.

Explanation:

There are

12 novels

8 plays

12 nonfiction.

He wants to include

5 novels

4 plays

2 nonfiction

The order in which the novels, plays and nonfictions are chosen is not important. So we use the combinations formula to solve this problem.

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

How many ways can he choose the books to put on the list?

Novels:

5 from a set of 12. So

`C_(12,5) = (12!)/(5!7!) = 792`

Plays:

4 from a set of 8. So

`C_(8,4) = (8!)/(4!4!) = 70`

Nonfiction:

2 from a set of 12

`C_(12,2) = (12!)/(2!10!) = 66`

Total:

Multiplication of novels, plays and nonfiction.

`T = 792*70*66 = 3,659,040`

There are 3,659,040 ways he can choose the books to put on the list.