Comparing sums of products of combinations

Question

I came across the following equation which seems to be true in all cases that I test:

Let integers $x \ge 5, y \ge 4$:

$${x \choose 2}{y \choose 2} = \sum_{i=1}^5(-1)^{i-1}{5\choose i}{{x-i}\choose2}{{y-i}\choose2}$$

It is not clear to me why this should be true in so many different cases.

Could someone help me to find a counterexample or help me understand why it is true.

Here are examples:

$$60={5\choose2}{4\choose2} = {5\choose1}{4\choose2}{3\choose2} - {5\choose2}{3\choose2}{2\choose2} = 90 - 30$$

$$100={5\choose2}{5\choose2}={5\choose1}{4\choose2}{4\choose2} - {5\choose2}{3\choose2}{3\choose2} + {5\choose3}{2\choose2}{2\choose2}= 180 - 90 + 10$$

Answer 1

Easier, I think, to show the equivalent formulation: $$\sum_{i=0}^5 (-1)^i\times \binom 5i\times \binom {x-i}2\times \binom {y-i}2=0$$ to see this use the identity $$\sum_{j=0}^n(-1)^j\times \binom nj\times P(j)=0$$ where $P(k)$ is any polynomial of degree $<n$, see this for a proof (it appears around the discussion of formula $(10)$). Your result follows once you note that, for all $(x,y)$, $\binom {x-i}2\times \binom {y-i}2$ is a polynomial of degree $4$ in $i$.

Answer 2

Note that we have $$\sum_{k=0}^nk^p\binom{n}{k}(-1)^k=0$$ for all $n\in\mathbb{N}$ and $p\in\{0,1,\dots,n-1\}$. This follows from taking the derivative of the identity $$\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n$$ and setting $x=-1$. In the case $n=5$ we get that $$\sum_{k=0}^5 k^p\binom{5}{k}(-1)^k=0$$ for $p\in\{0,1,2,3,4\}$. Then note that this is equivalent to saying that $$\sum_{k=0}^5 f(k)\binom{5}{k}(-1)^k=0$$ for any polynomial $f(k)$ with degree less than $5$. Now take $$f(k)=\binom{x-k}2\binom{y-k}2$$ and you should get the stated identity.