Recognize the series?

We can prove the claim that $A_{mn} = {m \choose n}$ is true for all $m, n \ge 0$ by induction.

1. For $m=n=0$ , $A_{00} = {0 \choose 0} = 1$ as defined in $A_{m0} \ \forall \ m \ge 0$ . Therefore, the claim is true for $m=n=0$ .

2. For $m=0$ and $n>0$ , $A_{00} = {0 \choose n} = 0$ as defined in $A_{0n} = 0$ for $n>0$ . Therefore, the claim is true for $m=0$ and $n>0$ .

3. Note that steps 1 and 2 prove that the claim is true for $m=0$ for all $n \ge 0$ . Now, assuming the claim $A_{mn} = {m \choose n}$ is true for $m$ , then:

$\begin{aligned} \quad A_{\color{#3D99F6}{m+1,n+1}} & = A_{m,n+1} + A_{m,n} \\ & = {m \choose n+1} + {m \choose n} & \small \color{#3D99F6}{\text{By Pascal's identity: }{n \choose k} + {n \choose k-1} = {n+1 \choose k}} \\ & = {\color{#3D99F6}{m+1} \choose \color{#3D99F6}{n+1}} \end{aligned}$ t

$\quad$ Therefore, the claim is true for $m+1$ and hence true for all $m \ge 0$ .

$\implies A_{15,7} = {15 \choose 7} = \boxed{6435}$

This series is just a fancy way of saying $A_{mn} = \binom{m}{n}$

So, $A_{15,7} = \binom{15}{7} = \boxed{6435}$