For non-negative integers $s$ and $r$, let $\binom{s}{r} = \begin{cases} \dfrac{s!}{r!\,(s-r)!} & \text{if } r \le s, \\[4pt] 0 & \text{if } r > s. \end{cases}$ For positive integers $m$ and $n$, let $g(m,n) = \sum_{p=0}^{m+n} \dfrac{f(m,n,p)}{\binom{n+p}{p}}$ where for any non-negative integer $p$, $f(m,n,p) = \sum_{i=0}^{p} \binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}.$ Then which of the following statements is/are TRUE?

$\binom{s}{r} = \dfrac{s!}{r!\,(s-r)!} \text{ if } r \le s, \text{ else } 0$ $f(m,n,p) = \sum_{i=0}^{p} \binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}$ $g(m,n) = \sum_{p=0}^{m+n} \dfrac{f(m,n,p)}{\binom{n+p}{p}}$ $\text{(1) } g(m,n) = g(n,m)$ $\text{(2) } g(m,n+1) = g(m+1,n)$ $\text{(3) } g(2m,2n) = 2g(m,n) \qquad \text{(4) } g(2m,2n) = (g(m,n))^{2}$ $\text{Goal: find a closed form for } g(m,n)$ $\text{Claim: } g(m,n) = 2^{m+n}$ $\binom{n+p}{p-i} = \binom{n+p}{n+i}$ $f(m,n,p) = \sum_{i=0}^{p} \binom{m}{i}\binom{n+i}{p}\binom{n+p}{n+i}$ $\dfrac{\binom{n+p}{n+i}}{\binom{n+p}{p}} = \dfrac{p!\,n!}{(n+i)!\,(p-i)!}$ $\binom{n+i}{p}\cdot\dfrac{p!\,n!}{(n+i)!\,(p-i)!} = \dfrac{n!}{(p-i)!\,(n-p+i)!}$ $\dfrac{n!}{(p-i)!\,(n-(p-i))!} = \binom{n}{p-i}$ $\dfrac{f(m,n,p)}{\binom{n+p}{p}} = \sum_{i=0}^{p} \binom{m}{i}\binom{n}{p-i}$ $\binom{n}{k}=0 \text{ if } k<0 \text{ or } k>n$ $\sum_{i=0}^{p}\binom{m}{i}\binom{n}{p-i} = \binom{m+n}{p}$ $\dfrac{f(m,n,p)}{\binom{n+p}{p}} = \binom{m+n}{p}$ $g(m,n) = \sum_{p=0}^{m+n} \binom{m+n}{p} = 2^{m+n}$ $g(1,1)=4,\quad g(1,2)=8,\quad g(2,2)=16$ $\text{Substitute } g(m,n)=2^{m+n} \text{ into each statement}$ $g(m,n)=2^{m+n}=2^{n+m}=g(n,m) \Rightarrow \text{(1) TRUE}$ $g(m,n+1)=2^{m+n+1}=g(m+1,n) \Rightarrow \text{(2) TRUE}$ $g(2m,2n)=2^{2m+2n}=4^{m+n} \neq 2^{m+n+1}=2g(m,n)$ $\Rightarrow \text{(3) FALSE}$ $g(2m,2n)=2^{2m+2n}=(2^{m+n})^{2}=(g(m,n))^{2}$ $\Rightarrow \text{(4) TRUE}$ $\text{Answer: (1), (2), (4) are TRUE}$