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kolezko [41]
3 years ago
13

An English teacher needs to pick 1111 books to put on his reading list for the next school year. He has narrowed down his choice

s 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?
Mathematics
1 answer:
forsale [732]3 years ago
4 0

Answer:

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

Step-by-step 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} = \frac{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} = \frac{12!}{5!7!} = 792

Plays:

4 from a set of 8. So

C_{8,4} = \frac{8!}{4!4!} = 70

Nonfiction:

2 from a set of 12

C_{12,2} = \frac{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.

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