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 $$\sum_{j=0}^k {n\choose j} \cdot{m\choose k -j} = {n+m\choose k}.$$

Hint: Suppose there are $n$ blue objects and $m$ red ones. How many ways can I choose $k$ objects if I don't care about the colors? Count that in two different ways. — lulu, Mar 13 '20 at 17:45
This is a well known problem, it is known as Vandermonde's identity. You can find this problem on this site. — J P, Mar 13 '20 at 17:48
https://math.stackexchange.com/questions/337923/how-to-prove-vandermondes-identity-sum-k-0n-binomrk-binommn-k — Jean-Claude Colette, Mar 13 '20 at 17:50
At least show some effort, it's like you are in a hurry and just want to give your solutions, which is extremely wrong. — , Mar 13 '20 at 19:20

score 0 · Answer 1 · answered Mar 13 '20 at 17:59

0

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

answered Mar 13 '20 at 17:59

Z Ahmed

score 0 · Answer 2 · answered Mar 13 '20 at 19:13

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

2 Answers2