TITLE: Sum involving binomial coefficients.
QUESTION [7 upvotes]: Prove that $${^{404}\mathrm C_4}-{^4\mathrm C_1}\cdot{^{303}\mathrm C_4}+{^4\mathrm C_2}\cdot{^{202}\mathrm C_4}-{^4\mathrm C_3}\cdot{^{101}\mathrm C_4} =(101)^4$$

I tried writing $101=102-1$, but couldn't move forward.

REPLY [6 votes]: The given expression here is the coefficient of $x^4$ in $P(x) := ((1+x)^{101}-1)^4$. You can see this by first expanding $(y-1)^4$, then putting $y = (1+x)^{101}$ and expanding it in terms of $x$.
But $P(x) =((1+101x+...)-1)^4 = (101x+...)^4 = 101^4\cdot x^4(1+...)^4$ where the dots represent higher powers of $x$.
Thus, the coefficient of $x^4$ in $P(x)$ is indeed $101^4$.