Find the sum
$$^{505}C_5-5\cdot{^{404}C_5}+10\cdot{^{303}C_5}-10\cdot{^{202}C_5}+5\cdot{^{101}C_5}-1$$
Critiques and other answers are always welcome!
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A generalization seems to be $$\sum_{k=0}^n {^{m(n-k)}C_n}{^n C_k}(-1)^k=\sum_{k=0}^n \binom{m(n-k)}{n}\binom{n}{k}(-1)^k=m^k.$$ It's not implausible that this has already been considered on this site. – Semiclassical Jan 20 '21 at 07:13
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@Semiclassical: See my answer for a combinatorial explanation for the general case. – user21820 Feb 02 '21 at 04:48
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We have to find the coefficient of $x^5$ in $$(1+x)^{505}-5.(1+x)^{404}+10.(1+x)^{303}-10.(1+x)^{202}+5.(1+x)^{101}-1=((1+x)^{101}-1)^5=({101}x+O(x^2))^5$$ Multinomially expand the final term and convince yourself that the coefficient of $x^5$ is $101^5$
DatBoi
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Nice method using coefficeint extractor that was my same idea too!+1 – Albus Dumbledore Jan 20 '21 at 06:55
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It is simply the inclusion-exclusion principle... Choosing $1$ from each of $k$ groups of $n$ is equivalent to choosing $k$ from all $k$ groups (combined) minus choosing $k$ from only $(k-1)$ groups plus choosing $k$ from only $(k-2)$ groups ...
user21820
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