Sum of the series in terms of coefficient of $x^m$ in some binomial expansion.

Question

It is written in my book that:
¹⁰C₀ײ⁰C_m + ¹⁰C₁ײ⁰C_m-1+ ... + ¹⁰C_mײ⁰C₀ = coefficient of $x^m$ in the expansion of $(1+x)^{10}(1+x)^{20}$. I'm in doubt how this happened.

Answer 1

This a particular case of Vandemonde's identity. By the Binomial Theorem $$(1+x)^N(1+x)^M=\left(\sum_{k=0}^N \binom{N}{k}x^k\right)\cdot \left(\sum_{j=0}^M \binom{M}{j}x^j\right)$$ So the coefficient of $x^m$ in the expansion of the above product is
$$\sum_{k+j=m} \binom{N}{k}\binom{M}{j}=\sum_{k=0}^m \binom{N}{k}\binom{M}{m-k}.$$

Answer 2

The binomial theorem gives us the following: $$(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k$$

So that $$(1+x)^{10} (1+x)^{20}=\left(\sum_{k=0}^{10}\binom{10}{k}x^k \right)\left(\sum_{r=0}^{20}\binom{20}{r}x^r\right)=\sum_{k=0}^{10}\sum_{r=0}^{20}\binom{10}{k}\binom{20}{r}x^{k+r}$$

Let $u=k+r$ then $$\sum_{k=0}^{10}\sum_{u=k}^{20+k}\binom{10}{k}\binom{20}{u-k}x^{u}=\sum_{u=0}^{30}\sum_{k=0}^{u}\binom{10}{k}\binom{20}{u-k}x^{u}$$

EDIT: Apparently, this result can be taken directly from Vandermonde's identity, but what I did is simply interchange the sums...