Following Cohen [1978], define $\genfrac{\langle}{\rangle}{0pt}{}{n}{r}$ to be $\binom{n+r-1}{r}$. Show that $$\genfrac{\langle}{\rangle}{0pt}{}{n}{r} = \genfrac{\langle}{\rangle}{0pt}{}{n}{r-1} + \genfrac{\langle}{\rangle}{0pt}{}{n-1}{r}$$ (a) using an algebraic proof

1. The left side of the equation: $$\genfrac{\langle}{\rangle}{0pt}{}{n}{r} = \binom{n+r-1}{r} = \frac{(n+r+1)!}{r!(n-1)!}$$

2. The right side of the equation: $$\genfrac{\langle}{\rangle}{0pt}{}{n}{r-1} + \genfrac{\langle}{\rangle}{0pt}{}{n-1}{r} = \binom{n+(r-1)-1}{r-1} + \binom{(n-1)+r-1}{r} = \binom{n+r-2}{r-1} + \binom{n+r-2}{r} \\ = \frac{(n+r-2)!}{(r-1)!((n+r-2)-(r-1))!} + \frac{(n+r-2)!}{r!(n-2)!} = \frac{(n+r-2)!}{(r-1)!(n-1)!} + \frac{(n+r-2)!}{r!(n-2)!} \\ = \frac{r(n+r-2)! + (n-1)(n+r-2)!}{r!(n-1)!} = \frac{(n+r-1)(n+r-2)!}{r!(n-1)!} = \frac{(n+r-1)!}{r!(n-1)!}$$

   Finally, we can see that both sides reduce to the same result.

2. using a combinatorial proof

TBD