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 about in our FAQ: 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.