Evaluate the sum $\sum_{k=1}^n {n \choose k-1} {n \choose k}$

Question

Evaluate the summation of: $$ \sum_{k=1}^n {n \choose k-1} {n \choose k} $$ I have been trying to solve this problem by translating C(n, k-1) into C(n, k), because I know that: $$ \sum_{k=1}^n {n \choose k}^2={2n \choose n} $$ However, I haven't been having any luck. Am I on the right track? What am I doing wrong?

Answer 1

$$\sum_{k=1}^n \binom{n}{k-1}\binom{n}{k} = \sum_{k=1}^{n}\binom{n}{n-k+1}\binom{n}{k}$$

This is the number of ways we can obtain $x^{n+1}$ in the product $(1+x^n)\cdot(1+x)^n$.

That is, your summation is the coefficient of $x^{n+1}$ in $(1+x)^n\cdot(1+x)^n$, that is the coefficient of $x^{n+1}$ is $(1+x)^{2n}$ which is $\binom{2n}{n+1}$.

Answer 2

Vandermonde's identity states $$ \sum_{k=0}^r \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r} $$ for $m, n, r \in \mathbb{N}_0$. Therefore, $$ \sum_{k=1}^{n} \binom{n}{k-1}\binom{n}{n-1 - (k-1)} = \binom{n+n}{n-1} $$

Answer 3

HINT $${n \choose k }= {n \choose {n-k}}$$

SOLUTION

$$\implies I = \sum_{k=1}^n{n \choose k-1}{n \choose n-k}$$ This can be seen to be equal to the coefficient of $x^{n-1}$ in the expansion of the product $(x+1)^n.(x+1)^n = (x+1)^{2n}$

Hence, answer is $I = {2n \choose n-1}$