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6 boys. Pick one (doesn't matter). After you pick one, you have 5 boys left from which you have to pick 2 more to make a team of 3. So C(5,2) = 10. Once the first team of 3 is picked, you're left 3 boys and only one way to pick C(3,3). So 10*1= 10.

I'd like to revive this question....I too would like to know why we divide by 2.

Thanks!

Last note of the question, part in the parathesis is the answer.

6C3, gives number of ways 6 people can be organized into group of 3 people.
f.e. group A and group B OR group B and group A, which is 20 in total, but since the order of groups does not matter group A and group B is the same as group B and group A, thus we have to devide by 2.

* Pick 3 boys for group A: Joe, John, James
--> This leaves 3 boys for group B: Mike, Mark, Mitch

* Pick 3 boys for group B: Mike, Mark, Mitch
--> This leaves 3 boys for group A: Joe, John, James

We see the two choices results in the same groups - this occurs because selection of the first group decided what boys would be in the second group. So half of the 6C3 results will be mirror images of each other --> divide the result by 2.

Re: Quick Combinations question... [#permalink]
03 Oct 2007, 11:33

plaguerabbit wrote:

How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

Explanations too, if you will

here the caveat is that the order of the groups do not matter. Therefore it is half of the total, since we only need to figure out how many groups of 3 we can create out of the six, but since there is no A or B group, there is only A and A group. so it should look like this 6!/3!2!3!

Re: Quick Combinations question... [#permalink]
03 Oct 2007, 11:46

plaguerabbit wrote:

How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

Explanations too, if you will

here the caveat is that the order of the groups do not matter. Therefore it is half of the total, since we only need to figure out how many groups of 3 we can create out of the six, but since there is no A or B group, there is only A and A group. so it should look like this 6!/3!2!3!

gmatclubot

Re: Quick Combinations question...
[#permalink]
03 Oct 2007, 11:46