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### Committee of Six

```
Date: 10/07/1999 at 13:45:54
From: Samantha
Subject: Grouping/order

A club has 8 male and 8 female members. The club is choosing a
committee of 6 members. The committee must have 3 male and 3 female
members. How many different committees can be chosen?

Can you give me a systematic method to approach questions like this? I
homeschool my kids and would really appreciate it.

Thanks,
Samantha
```

```
Date: 10/07/1999 at 16:55:14
From: Doctor Peterson
Subject: Re: Grouping/order

Hi, Samantha.

This problem involves what we call "combinations," which you can learn

http://mathforum.org/dr.math/faq/faq.comb.perm.html

Since this is a fairly complicated problem in this area, I have to
assume that you have already covered some aspects of combinations and
permutations; if I go too fast, just write back and tell me where to
go slower.

Picture the committee we are trying to fill as six chairs, three blue
and three pink. (Or just label them "male" and "female" if that's too
cute for you.) To choose a committee, we have to first choose three
male members, then choose three female members. How many ways are
there to fill the blue chairs? We can choose any of 8 people for the
first, any of the 7 who are left for the second, and any of the 6 who
remain for the third. This gives

8 * 7 * 6

ways to fill the three chairs. But it doesn't matter which man is in
which of the blue chairs; there are

3 * 2 * 1

ways to rearrange our three men, so our count of 8*7*6 includes
3*2*1 = 6 "different" committees for each possible choice that is
really different. The actual number of different groups of three men
will be

8 * 7 * 6
--------- = 56
3 * 2 * 1

This is called the number of combinations of 8 taken 3 at a time, or

C
8 3

If you know this formula, you don't need to do all the thinking I've
done so far.

Now once we've chosen the men, we do the same thing to choose the
women; there are 56 ways to do that, too. But for each choice of the
men, we can make any of the 56 choices of women, so the total number
of different committees we can choose is

56 * 56 = 3136

There are several ways in which I approached this systematically. One
is to take things one step at a time: first choice, second choice,
etc. Another is to picture the problem in a way that makes the
counting as concrete as possible: chairs, not just lists. Then in each
case I recognized when I can use multiplication; if the second choice
is independent of the first (for each A, I can do any of B), then I
can multiply. In working out the combinations, I have a different set
of people for the second choice depending on whom I chose for the
first chair, but it's always the same number, so I can still multiply.

Finally, I always have to think carefully about whether order matters.
By making things concrete and taking it one step at a time, I may have
overdone it, and made the order matter where it shouldn't have. In
choosing the men, I had to correct for this, because the "blue" chairs
are all the same, and there's no real difference between having John
in chair 1 and Tim in chair 2, and the other way around. In combining
the choices of men and women, there was no problem, because the "blue"
and "pink" chairs are different: whatever committee I've chosen, I can
herd the men into one set of chairs and the women into the other.

I hope that helps you out. Feel free to write back until you feel
comfortable with this.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations

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