For nonnegative integers s and r, let
$$\left(\begin{array}{c}s\\ r\end{array}\right) = \begin{cases}\frac{s!}{r!(s - r)!} & if r \leq s\\0 & if r > s\end{cases}$$
For positive integers 𝑚 and 𝑛, let
$$g(m, n) = \sum_{p=0}^{m+n}\frac{f(m, n, p)}{\left(\begin{array}{c}n+p\\ p\end{array}\right)}$$
where for any nonnegative integer 𝑝,
$$f(m, n, p) = \sum_{i=0}^{p}\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}n+i\\ p\end{array}\right)\left(\begin{array}{c}p+n\\ p - i\end{array}\right)$$
Then which of the following statements is/are TRUE?

We begin with the given definitions. For non-negative integers $$s,r$$

$$\left(\begin{array}{c}s\\ r\end{array}\right)= \begin{cases} \dfrac{s!}{r!(s-r)!}, & r\le s\\[4pt] 0, & r\gt s \end{cases}$$

For positive integers $$m,n$$ we have

$$g(m,n)=\sum_{p=0}^{m+n}\dfrac{f(m,n,p)}{\left(\begin{array}{c}n+p\\ p\end{array}\right)}$$

where

$$f(m,n,p)=\sum_{i=0}^{p}\left(\begin{array}{c}m\\ i\end{array}\right) \left(\begin{array}{c}n+i\\ p\end{array}\right) \left(\begin{array}{c}p+n\\ p-i\end{array}\right).$$

Our objective is to evaluate $$g(m,n)$$ and then verify the four statements.

Step 1 : Simplifying the inner summand

Rewrite the third binomial as $$\left(\begin{array}{c}p+n\\ p-i\end{array}\right)= \left(\begin{array}{c}p+n\\ n+i\end{array}\right).$$

Divide each term of the inner sum by the denominator of $$g(m,n)$$:

$$T_{i,p}=\dfrac {\displaystyle\left(\begin{array}{c}m\\ i\end{array}\right) \left(\begin{array}{c}n+i\\ p\end{array}\right) \left(\begin{array}{c}p+n\\ p-i\end{array}\right)} {\displaystyle\left(\begin{array}{c}n+p\\ p\end{array}\right)}.$$

Using factorial definitions,

$$\dfrac{\left(\begin{array}{c}p+n\\ p-i\end{array}\right)} {\left(\begin{array}{c}n+p\\ p\end{array}\right)} =\dfrac{(n+p)!}{(p-i)!\,(n+i)!}\cdot \dfrac{p!\,n!}{(n+p)!} =\dfrac{p!\,n!}{(p-i)!\,(n+i)!}.$$ Also $$\left(\begin{array}{c}n+i\\ p\end{array}\right)= \dfrac{(n+i)!}{p!\,(n+i-p)!}.$$

Therefore

$$T_{i,p}= \left(\begin{array}{c}m\\ i\end{array}\right) \dfrac{(n+i)!}{p!\,(n+i-p)!}\; \dfrac{p!\,n!}{(p-i)!\,(n+i)!} = \left(\begin{array}{c}m\\ i\end{array}\right) \dfrac{n!}{(p-i)!\,(n+i-p)!}.$$

The denominator inside equals that of the binomial $$\left(\begin{array}{c}n\\ p-i\end{array}\right)$$, hence

$$T_{i,p}= \left(\begin{array}{c}m\\ i\end{array}\right) \left(\begin{array}{c}n\\ p-i\end{array}\right).$$

Step 2 : Rewriting $$g(m,n)$$ with the simplified term

Plugging $$T_{i,p}$$ into the definition of $$g(m,n)$$:

$$g(m,n)=\sum_{p=0}^{m+n}\;\sum_{i=0}^{p} \left(\begin{array}{c}m\\ i\end{array}\right) \left(\begin{array}{c}n\\ p-i\end{array}\right).$$

Step 3 : Changing the order of summation

Put $$j=p-i$$ so that $$j\ge 0$$ and $$i+j=p$$. The constraints $$0\le i\le m$$ and $$0\le j\le n$$ automatically guarantee $$0\le p=i+j\le m+n$$. Hence every ordered pair $$(i,j)$$ with $$0\le i\le m,\,0\le j\le n$$ appears exactly once.

Therefore the double sum factorises:

$$g(m,n)=\sum_{i=0}^{m}\left(\begin{array}{c}m\\ i\end{array}\right) \sum_{j=0}^{n}\left(\begin{array}{c}n\\ j\end{array}\right) =\Bigg(\sum_{i=0}^{m}\left(\begin{array}{c}m\\ i\end{array}\right)\Bigg) \Bigg(\sum_{j=0}^{n}\left(\begin{array}{c}n\\ j\end{array}\right)\Bigg).$$

Step 4 : Using the binomial theorem

The well-known identity $$\sum_{k=0}^{r}\left(\begin{array}{c}r\\ k\end{array}\right)=2^{r}$$ gives

$$g(m,n)=2^{m}\,2^{n}=2^{\,m+n}.$$

Step 5 : Verifying the statements

A. $$g(m,n)=2^{\,m+n}=2^{\,n+m}=g(n,m).$$ True.

B. $$g(m,n+1)=2^{\,m+n+1}=2^{\,m+1+n}=g(m+1,n).$$ True.

C. $$g(2m,2n)=2^{\,2m+2n}=4^{\,m+n},\quad 2\,g(m,n)=2^{\,m+n+1}.$$ For $$m+n\gt 1$$ the two sides are unequal; hence the relation is not generally valid. False.

D. $$g(2m,2n)=2^{\,2(m+n)}=\big(2^{\,m+n}\big)^{2}=\big(g(m,n)\big)^{2}.$$ True.

Final result
The correct statements are Option A, Option B and Option D.