How to simpify the following equation involving binomial coefficients? How can one simplify this equation:
$$
\sum_{k=0}^{n-1}\binom{n}{k}\binom{n}{k+1}
$$
 A: Algebraic argument:
$$(1+x)^n = \sum_{k=0}^{n} \dbinom{n}k x^k$$
$$(1+x)^n = \sum_{k=0}^{n} \dbinom{n}k x^{n-k}$$
Multiplying the above two we get that
$$(1+x)^{2n} = \sum_{j=0}^n \sum_{k=0}^n \dbinom{n}j \dbinom{n}k x^{n-k+j}$$
Hence, the coefficient of $x^{n+1}$ in the product of the above two is (i.e. $j=k+1$)
$$\sum_{k=0}^{n-1} \dbinom{n}{k+1} \dbinom{n}k$$
The coefficient of $x^{n+1}$ in $(1+x)^{2n}$ is
$$\dbinom{2n}{n+1}$$
Hence,
$$\sum_{k=0}^{n-1} \dbinom{n}{k+1} \dbinom{n}k = \dbinom{2n}{n+1}$$

Combinatorial argument:
Consider a bag with $n$ red balls and another bag with $n$ blue balls. Number of ways of choosing a total of $n+1$ balls is given by $$\dbinom{2n}{n+1}$$
Another way to count the same thing is if we select $k+1$ red balls from $n$ red balls, then we need to select $n-k$ blue balls from $n$ blue balls (equivalently we need to reject $k$ blue balls from $n$ blue balls). Hence, number of ways to do this is $\dbinom{n}{k+1} \dbinom{n}k$. To take all possible cases into account, we need to run $k$ from $0$ to $n-1$. Hence, we get the total number of ways is $$\sum_{k=0}^{n-1} \dbinom{n}k \dbinom{n}{k+1}$$
