How to prove that $\sum_{j=0}^k {n\choose j} \cdot{m\choose k -j} = {n+m\choose k}$

Question

I am a high school student and I need help proving that $$\sum_{j=0}^k {n\choose j} \cdot{m\choose k -j} = {n+m\choose k}.$$

Answer 1

$$S=\sum_{j=0}^{k} {k \choose j} {m \choose k-j}=$$ Coefficient of $x^{j+k-j}=x^k$ in $$(1+x)^{n+m}= {n+m \choose k}$$

Answer 2

Lets say that you have a committee of $n+m$ people, and $n$ are males and $m$ are females.

Now in how many ways can you choose $k$ people

$\binom {n+m}{k}$

Now another way to count this would be to consider all possible combinations of $j$ males and $k-j$ females and this can be done in

$\sum_{j=0}^k \binom{n}{j} \cdot \binom {m}{k-j}$

Hence both sides are equal