I am trying to find a nice way to solve for $$\sum_{a,b,c\ge 1, a+b+c=9}\frac{9!}{a!b!c!}=18150$$ I managed to solved it (and verified by computer) by doing manually (on paper) on $7$ cases and got a "nice" answer of $216+1512+3024+2268+1890+7560+1680=18150=2\cdot 3\cdot5^2 \cdot11^2$.
2026-03-30 10:40:08.1774867208
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Finding a nice solution to $\sum_{a,b,c\ge 1, a+b+c=9}\frac{9!}{a!b!c!}=18150$
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This is the number of surjections from a 9-set onto a 3-set. More generally, the number of surjections from an $n$-set onto a $k$-set involves Stirling numbers of the second kind: https://oeis.org/A019538
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Given $a,b,c$, each summand can be thought of as the number of ways to fill $9$ slots with $a$ blue marbles, $b$ red marbles, and $c$ yellow marbles. Therefore, the sum is the total number of ways to fill $9$ slots with blue, red, and yellow marbles assuming at least one of each. You can solve this with the inclusion/exclusion principle as $$3^9-3\cdot2^9+3\cdot1^9=18150$$