3. The following questions involve permutations and combinations. You may leave

Question

3. The following questions involve permutations and combinations. You may leave your answer in terms of factorials. (a) (2 points) How many ways can you pick 4 out of 8 people to come to a dinner party? (b) (3 points) There are 10 players on a basketball team. The first five starting players are introduced one at a time. In how many different ways can they be introduced? IO oxax (c) (5 points) How many batting orders are possible for a team of nine baseball players in which the pitcher always bats last and the first baseman bats in either the third or fourth spot? (d) (5 points) Fred has 12 books. He plans to give three books to Jill and four books to Jack. How many different ways are there for him to do this? 12,4

Explanation / Answer

(a) Since order is not important, the number of ways to pick 4 people out of 8

= 8C4

= 8! / (4! * (8 - 4)!)

= 8! / (4! * 4!)

= (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1)

= 70.

(b) In this case, the order is important. There are 10 players out of whom we need to arrange 5 players and this can be done in 10P5 ways.

10P5 = 10! / (10 - 5)!

= 10! / 5!

= 10 * 9 * 8 * 7 * 6

= 30240.

(c) Since the pitcher bats last, we need to only worry about the remaining 8 positions.

There are two positions for the first baseman and the remaining 7 players take the 7 remaining positions.

Total number of batting orders

= 2 * 7!

= 2 * 5040

= 10080.

(d) The three books to be given to Jill can be chosen out of 12 in 12C3 ways and the four books to be given to Jack can be chosen out of the remaining 9 in 9C4 ways.

Total number of ways to give the books

= 12C3 * 9C4

= (12 * 11 * 10) / (3 * 2 * 1) * (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1)

= 27720.

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