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If you want to divide a team of 10 people into teams A, B, and C of sizes 3,5, and 2, how many divisions are possible?

If you want to divide them into just teams of sizes 3,5, and 2, how many arrangements are possible?

I know that you use multinomial coefficients such that for part 1, the number of divisions is 10!/(3!5!2!) and for part 2, the number of divisions is 10!/(3!5!2!)/2!. I don't know why this is the case intuitively. Also, I can understand the formula for combinations as (n choose k) = (n*n-1*..n-k+1)/(k!) more clearly than (n!)/(k!)(n-k)!. I can't seem to interpret the second form of of the formula for combinations(or multinomials).

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    Basically you ask how many total divisions are possible, and then you divide by how many times you counted each option. Say the team (123)(45678)(9,10) is the same as (132)(57648)(10,9). So you divide by the number of permutations of each group, which is $k!$ and $(n-k)!$ for the binomial case.2011-09-26
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    How can the same question have different answers? (It is true that $3-4-3$ has different shape than $3-5-2$.)2011-09-26

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