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First of all, which question are you referring to? There was the original question and then a subsequent question that was modified. The second question was 3 teams, First = 2 people, Second = 2 people and the 3rd = 4 people (total of 8) people.

Can you tell us which question you're talking about so we can explain ourselves?

rino wrote:

I dont see why shouldn't it be a simple 8c2.

i think u are all wrong

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

The first team of is a combination of 2 things taken from 8, or C(8,2), then the next is a team of 2 taken from 6 available choices, so C(6,2), and the final team you state is a team of 4 taken from 4 available choices C(4,4). We have a total of 3 teams and we do not want to treat the order of the teams differently so we divide that by 3!. Dividing by 3! removes the number of ways that each team can be ordered because unless we do that, we count the same teams, but in different order as a different combination, which is not what the question asks for. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

The first team of is a combination of 2 things taken from 8, or C(8,2), then the next is a team of 2 taken from 6 available choices, so C(6,2), and the final team you state is a team of 4 taken from 4 available choices C(4,4). We have a total of 3 teams and we do not want to treat the order of the teams differently so we divide that by 3!. Dividing by 3! removes the number of ways that each team can be ordered because unless we do that, we count the same teams, but in different order as a different combination, which is not what the question asks for.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...