# Sum calculation with binomial coefficients

Knowing that $$\sum_{k} \binom{a+b}{a+k} \binom{b+c}{b+k} \binom{c + a}{c + k} (-1)^k = \frac{(a + b + c)!}{a!b!c!}$$ where $a,b,c ∈ \mathbb Z$ solve the following sum $$\sum_{k} \binom{2a}{a+k} \binom{2b}{b+k} \binom{2c}{c + k} (-1)^k$$

We have $$\sum_k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}(-1)^k =\sum_k\frac{(a+b)!(b+c)!(c+a)!(-1)^k}{(a+k)!(b-k)!(b+k)!(c-k)!(c+k)!(a-k)!} \\ =(a+b)!(b+c)!(c+a)!\sum_k\frac{(-1)^k}{(a+k)!(b-k)!(b+k)!(c-k)!(c+k)!(a-k)!}$$ and $$\sum_k\binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k}(-1)^k =\sum_k\frac{(2a)!(2b)!(2c)!(-1)^k}{(a+k)!(a-k)!(b+k)!(b-k)!(c+k)!(c-k)!} \\ =(2a)!(2b)!(2c)!\sum_k\frac{(-1)^k}{(a+k)!(a-k)!(b+k)!(b-k)!(c+k)!(c-k)!}$$ The term behind the sum is the same in both cases, thus we have $$\sum_k\binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k}(-1)^k \\ = \frac{(2a)!(2b)!(2c)!}{(a+b)!(b+c)!(c+a)!}\sum_k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}(-1)^k \\ = \frac{(2a)!(2b)!(2c)!(a+b+c)!}{(a+b)!(b+c)!(c+a)!a!b!c!}$$