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How to prove, that $S(n, n-3) = \frac{n(n-1)(n-2)^2(n-3)^2}{48}$.

I want to go the combinatoric way. I've already seen this question ( Find a closed formula for the stirling number $S(n, n-3)$ for n $\ge$ $3$ ) but I don't understand it.

It's clear, that there are...

  1. case: 4 elements in the first box
  2. case: 3 elements in the first box, 2 in the second
  3. case: 2 elements in the first box, 2 in the second and 2 in the third.

It's clear to me that 1. case is $\binom{n}{4}$, but why is for example the 3. case $\frac{1}{6} \binom{n}{2} \binom{n-2}{2} \binom{n-4}{2}$ ?? Where does the $\frac{1}{6}$ come from?

Quotenbanane
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2 Answers2

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The boxes are unmarked, so (12)(34)(56) is the same as (34)(12)56). So, you have to divide by $6=3!.$

Igor Rivin
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${n \brace m}$ represents the number of ways to partition the set $\{1,2,\cdots,n\}$ into $m$ subsets.
Now you have to pay attention to the fact that the order of the sub-sets does not count, i.e. that the sub-sets are themselves arranged into a set.
Then when you have $2, 3, 4,\ldots $ subsets with the same size, you have to divide by the factorial of their number. Same, if you have $n_a$ subsets with size $a$, $n_b$ with size $b$, etc. you shall divide by $n_a! n_b! \ldots$.

So the $6 = 3!$ comes from the correction on having $3$ subsets of same size.

G Cab
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