# Help me solve this problem

1. Oct 4, 2007

### Diane Olive

Given a club consisting of six men and seven women

a. In how many ways can we select a committee of five persons?

b. In how many ways can we select a committee of five persons with 2 ment and 3 women?

2. Oct 4, 2007

### arildno

This is a very difficult problem.
None here at these forums will be able to solve it, unless YOU POST WHAT YOU HAVE DONE SO FAR!!

Last edited by a moderator: Oct 4, 2007
3. Oct 4, 2007

### Avodyne

Actually, it's not that difficult. But arildno is right that you must show an attempt ...

4. Oct 4, 2007

### Dick

I guess people not showing work is making arildno crabby. Try to avoid this.

5. Oct 5, 2007

### HallsofIvy

Staff Emeritus
Absolutely, you wouldn't like arildno when he is angry!

6. Oct 5, 2007

### Diane Olive

I got it

I went to a study help session last night and was able to get the answer. I would really like to have help in understanding combination, permutation, and other related problems.

7. Oct 5, 2007

### Diane Olive

Here is what we came up with last night:

Given a club consisting of six men and seven women
a. In how many ways can we select a committee of five persons?

This is a fundamental counting principle: 13 X 12 X 11 X 10 X 9 = 154,440

I will find out today if I was right or not.

b. In how many ways can we select a committee of 5 persons with two men and 3 women?

This is 7 choose 3 times 6 choose 2 = 7! divided by 3! X 4! times 6! divided by 2! times 4!
when you simplify I get = 525

8. Oct 5, 2007

### arildno

Completely correct!
The binomial, $$\binom{7}{3}\equiv\frac{7!}{3!4!}$$ tells us that
i) There are 7! permutations of 7 women. When these women are divided into two groups, (those 3 to be in the committee and those 4 who don't), then there are 3!*4! different permutations of a particular choice.

Since permutations of a particular choice is irrelevant, we have to divide 7! with 3!4! to get the number of DISTINCT 3-committe groups of women. (i.e, 35)
The number of distinct women groups of 3 is then multiplied with the number of distinct men groups of 2 (15), yielding a total of 525 possible committee choices.

Last edited: Oct 5, 2007
9. Oct 5, 2007

### Dick

Part a) didn't go so well though. Just as arildno explained, the number of ways of picking 3 from 7 is $$\binom{7}{3}$$. a) asks you to pick 5 from 13, why isn't the answer $$\binom{13}{5}$$?

10. Oct 5, 2007

### arildno

Oops, you are right!

11. Oct 20, 2007

### KenKEN Stone

it's not a difficult problem!!!!!