Q&A #19440


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From: Gail (for Teacher2Teacher Service)
Date: Mar 30, 2008 at 08:30:02
Subject: Re: Combinations

Hi Sandra, Let's try an easier problem first. How many committees consisting of 1 person can be formed from 3 people? Just 3, right? If the people are a, b, and c, you can have a, b, or c as committees. What about forming two-person committees? Then you could have ab, ac, ba, bc, ca or cb as committees, right? That is 6 committees. But wait a minute… wouldn’t the committee of ab be the same as the committee of ba? So you started with 6 options, but discarded 3 of them, and really you only have ab, ac, or bc, three committees. Let’s try it again, but this time, let’s use four people: a, b, c, and d. Two person committees: ab, ac, ad, bc, bd, cd (we discarded ba, ca, cb, da, db, and dc). So you started with 12 options, and threw out 6 of them. Here is a picture of what is happening: AB ba ca da AC BC cb db AD BD CD dc The ones we are using are in capital letters, and the ones that are duplicated are lower case. Now, what about three person committees? If you start with three people, a, b, and c, there is only one committee, abc. If you have four people to start with (a, b, c and d), the committees might be abc, abd, or acd or bcd. All of the other combinations would be duplicates of the ones listed. Here is a picture of what is happening: ABC bac cab dab ABD bad cad dac acb bca cba dba ACD BCD cbd dbc adb bda cda dca adc bdc cdb dcv What if you start with 5 people in your group (a, b, c, d and e), and want to choose 3-person committee? abc, abd, abe, acd, ace, ade, bcd, bce, bde, and cde are left if you discard all the duplicates. I found the following sites that could also be of help: http://mathforum.org/dr.math/faq/faq.mcdonalds.html http://mathforum.org/dr.math/faq/faq.comb.perm.html http://mathforum.org/dr.math/faq/faq.pascal.triangle.html I hope this is helpful. -Gail, for the T2T service

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