How many partitions of $12$ are there that have at least four parts, such that the largest, second-largest, third-largest, and fourth-largest parts are respectively greater than or equal to $4,3,2,1$?
For example, $12=4+4+2+2$ works.
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1There aren't that many. Simply listing them out works. Have you tried that? – JimmyK4542 Jul 18 '14 at 00:57
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1Oh yeah, I've got this problem but I wanted to know if there was a different way. – asdf Jul 18 '14 at 00:58
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1OK, is $\binom52+1+1=12$ the right answer? – bof Jul 18 '14 at 01:08
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Yes! How did you get that expression? – asdf Jul 18 '14 at 01:12
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1$\binom52$ ways to increase two of the five "piles" $4,3,2,1,0$ by $1$, plus two solutions that can't be obtained in this way, namely $6+3+2+1$ and $4+3+2+1+1+1$. – bof Jul 18 '14 at 01:16