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combination or permutation?

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by billzhao » Fri Feb 06, 2009 6:20 am
A group of 8 friends want to play double tennis. How many different ways can the group be divided into 4 teams of 2 people?

Logic thinking: 1st group: C(2,8); 2nd group: C(2,6); 3rd group: C(2,4) and 4th group: C(2,2). So the total combination is: C(2,8)*C(2,6)*C(2,4)*C(2,2) = 2,520

Is it correct? I feels that the number of combination should be less that than that figure.
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Re: combination or permutation?

by betamax » Fri Feb 06, 2009 10:01 am
billzhao wrote:A group of 8 friends want to play double tennis. How many different ways can the group be divided into 4 teams of 2 people?

Logic thinking: 1st group: C(2,8); 2nd group: C(2,6); 3rd group: C(2,4) and 4th group: C(2,2). So the total combination is: C(2,8)*C(2,6)*C(2,4)*C(2,2) = 2,520

Is it correct? I feels that the number of combination should be less that than that figure.
That doesn't seem to be right. C(8,2) gives you the number of ways you can choose 2 people from a group of 8. So you have 28 possible teams of 2. If you want to create 4 teams of 2, then you need to choose from the 28 teams you have already created. So C(28,4) which is a number much larger than you have there.

Hmm..my response is fatally wrong! :roll:
Last edited by betamax on Fri Feb 06, 2009 10:36 am, edited 2 times in total.
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by Ian Stewart » Fri Feb 06, 2009 10:01 am
The answer you give would be correct if the question asked "From 8 people, in how many ways can you choose one team of two to go to the Olympics, one team of two to go to Wimbledon, one team of two to go to the Australian Open and one team of two to go to the French Open?" That is, in your solution, you're assuming that the order of the teams is important - you'd be counting (A, B), (C, D), (E, F), (G, H) as being different from (C, D), (A, B), (G, H), (E, F), for example. If the order of the teams is not important, as I think we should assume given how you phrased the question, we need to divide by 4!, the number of different orders we could make of the four teams. So the answer ought to be:

8C2 * 6C2 * 4C2 * 2C2 / 4! = 105
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by billzhao » Fri Feb 06, 2009 4:55 pm
I got it. Thanks Ian Stewart!
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