Prove identity $\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$

Question

I found hard (for me at least) example, in fact it just would be nice to prove. $$\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$$

Answer 1

Suppose there are $n$ different red balls and $n$ different blue balls. In how many ways can we choose n balls from these $2n$ balls?

The answer can be directly written to be $\binom{2n}{n}$.

Now, think differently.

1. We can choose no red balls and all blue balls. We can do this in ${\binom{n}{0}}{\binom{n}{n}}$ ways.

2. We can choose $1$ red ball and $n-1$ blue balls. So total ${\binom{n}{1}}{\binom{n}{n-1}}$

.... This way we can choose $k$ red balls and $n-k$ blue balls. So, total $\binom{n}{k}\binom{n}{n-k}$ ways.

So, combining these all cases we get that there are total $\sum_{k=0}^{n} {\binom{n}{k}}{\binom{n}{n-k}}$ ways. And definitely this must be equal to $\binom{2n}{n}$. As both says the number of values of number of ways of this 'picking up $n$' game.

So, $\binom{2n}{n}=\sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$.

Answer 2

Hint $$(1+x)^{2n}=(1+x)^n (1+x)^n$$

$$\sum_{k=0}^{2n} \binom{2n}{k}x^k=\left(\sum_{k=0}^{n} \binom{n}{k}x^k \right)\left(\sum_{k=0}^{n} \binom{n}{k}x^k \right)$$

Now identify the coefficients of $x^n$.

Answer 3

Using Binomial theorem : $$(1+x)^n=\sum_{k=0}^n{\binom{n}{k}x^k}=\sum_{k=0}^n{\binom{n}{n-k}x^{n-k}}$$ Then : $$\sum_{k=0}^n\sum_{j=0}^n{\binom{n}{k}\binom{n}{n-j}x^{n-j+k}}=(1+x)^{2n}=\sum_{k=0}^{2n}{\binom{2n}{k}x^k}$$ From another hand the coefficient of $x^n$ in $\sum_{k=0}^n\sum_{j=0}^n{\binom{n}{k}\binom{n}{n-j}x^{n-j+k}}$ is $\sum_{k=0}^n{\binom{n}{k}\binom{n}{n-k}}$ which is equal to $\binom{2n}{n}$ usinig the other expression.