Can anyone help me with these word problems?

Question

-A baseball stadium has four entrance gates and nine exits. In how many ways may two men enter together, but leave by different exits?

-In how many ways can a bridge hand of 13 cards be chosen from a deck of 52 cards?

-In how many ways can a person get bridge hand consisting of two aces, one king, one queen, three jacks and the six others cards ten or less?

-From a group of 15 people, how many committees can be formed consisting of 2, 3, or four people?

Answer 1

This platform does not allow me to use the normal notation for the binomial coefficient "n chose k",

so I will use this notation: C(n,k)

- "A baseball stadium has four entrance gates and nine exits. In how many ways may two men enter together, but leave by different exits?"

The two men can enter together in C(4,1) ways (4 choices of an entrance).

They can exit in C(9,2) ways (a pair of exits chosen from 9 exits).

So the final answer is

C(4,1) x C(9,2) = 4 x 9 x 8 / 2 = 144

- "In how many ways can a bridge hand of 13 cards be chosen from a deck of 52 cards?"

By definition of the binomial coefficient, the answer is

C(52,13) = 52!/(13! x 39!) = 635,013,559,600

-"In how many ways can a person get bridge hand consisting of two aces, one king, one queen, three jacks and the six others cards ten or less?"

The two aces involve choosing 2 cards out 4: C(4,2) = 6

The one king involves choosing 1 card out 4: C(4,1) = 4

The one queen involves choosing 1 card out 4: C(4,1) = 4

The three jacks involve choosing 3 cards out 4: C(4,3) = 4

The remaining six cards involve involve choosing 6 cards out of 36: C(36,6) = 1,947,792

So the answer is the product of the five numbers:

6 x 4 x 4 x 4 x 1,947,792 = 747,952,128

-"From a group of 15 people, how many committees can be formed consisting of 2, 3, or four people?"

The answers are respectively

C(15,2) = 105

C(15,3) = 455

C(15,3) = 1365