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).
Best Answer
Think about it as if you have $n$ unique items that you want to divide between $m$ bins/groups of people/etc. - each size $k_1,k_2,\cdots,k_m$. Obviously the first gets $k_1$ items in $\binom{n}{k_1}$ number of ways. Obviously, the second one has to select from ($n-k_1$) items, which is $\binom{n-k_1}{k_2}$ and so on. Now take the product of these binomial coefficients and cancel out some terms to obtain $\binom{n}{k_1 k_2 \cdots k_m} = \frac{n!}{k_1! k_2! \cdots k_m!}$