Evaluation of Combinatorial Function Sum Identity

To determine the correct options, we simplify the expression for $g(m,n)$ step-by-step.

Step 1: Simplify the term inside the summation

The given function is:

$$f(m,n,p) = \sum_{i=0}^{p} \begin{pmatrix} m \\ i \end{pmatrix} \begin{pmatrix} n+i \\ p \end{pmatrix} \begin{pmatrix} p+n \\ p-i \end{pmatrix}$$

We consider the ratio:

$$\frac{f(m,n,p)}{\begin{pmatrix} n+p \\ p \end{pmatrix}} = \sum_{i=0}^{p} \begin{pmatrix} m \\ i \end{pmatrix} \frac{\begin{pmatrix} n+i \\ p \end{pmatrix} \begin{pmatrix} p+n \\ p-i \end{pmatrix}}{\begin{pmatrix} n+p \\ p \end{pmatrix}}$$

Let us expand the binomial coefficients in terms of factorials:

$$\begin{pmatrix} n+i \\ p \end{pmatrix} = \frac{(n+i)!}{p!(n+i-p)!}$$ $$\begin{pmatrix} p+n \\ p-i \end{pmatrix} = \frac{(p+n)!}{(p-i)!(n+i)!}$$ $$\begin{pmatrix} n+p \\ p \end{pmatrix} = \frac{(n+p)!}{p!n!}$$

Multiplying the numerator terms:

$$\begin{pmatrix} n+i \\ p \end{pmatrix} \begin{pmatrix} p+n \\ p-i \end{pmatrix} = \frac{(n+i)!}{p!(n+i-p)!} \cdot \frac{(p+n)!}{(p-i)!(n+i)!} = \frac{(p+n)!}{p!(n+i-p)!(p-i)!}$$

Now, dividing by $\begin{pmatrix} n+p \\ p \end{pmatrix}$:

$$\frac{\begin{pmatrix} n+i \\ p \end{pmatrix} \begin{pmatrix} p+n \\ p-i \end{pmatrix}}{\begin{pmatrix} n+p \\ p \end{pmatrix}} = \frac{\frac{(p+n)!}{p!(n+i-p)!(p-i)!}}{\frac{(n+p)!}{p!n!}} = \frac{n!}{(n+i-p)!(p-i)!} = \begin{pmatrix} n \\ p-i \end{pmatrix}$$

Thus, the expression simplifies to:

$$\frac{f(m,n,p)}{\begin{pmatrix} n+p \\ p \end{pmatrix}} = \sum_{i=0}^{p} \begin{pmatrix} m \\ i \end{pmatrix} \begin{pmatrix} n \\ p-i \end{pmatrix}$$

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Step 2: Evaluate $g(m,n)$ using Vandermonde's Identity

By Vandermonde's Identity, the convolution sum is:

$$\sum_{i=0}^{p} \begin{pmatrix} m \\ i \end{pmatrix} \begin{pmatrix} n \\ p-i \end{pmatrix} = \begin{pmatrix} m+n \\ p \end{pmatrix}$$

Substitute this result into $g(m,n)$:

$$g(m,n) = \sum_{p=0}^{m+n} \begin{pmatrix} m+n \\ p \end{pmatrix}$$

Using the sum of binomial coefficients formula $\sum_{k=0}^{N} \begin{pmatrix} N \\ k \end{pmatrix} = 2^N$, we get: $g(m,n) = 2^{m+n}$

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Step 3: Verification of the Statements

• Option (A): $g(m,n) = 2^{m+n}$ $g(n,m) = 2^{n+m} = 2^{m+n}$ Hence, $g(m,n) = g(n,m)$ for all positive integers $m, n$. (Statement A is TRUE)

• Option (B): $g(m,n+1) = 2^{m+(n+1)} = 2^{m+n+1}$ $g(m+1,n) = 2^{(m+1)+n} = 2^{m+n+1}$ Hence, $g(m,n+1) = g(m+1,n)$ for all positive integers $m, n$. (Statement B is TRUE)

• Option (C): $g(2m,2n) = 2^{2m+2n} = 2^{2(m+n)}$ $2 g(m,n) = 2 \cdot 2^{m+n} = 2^{m+n+1}$ Since $m, n \ge 1 \implies m+n \ge 2$, $2(m+n) \neq m+n+1$. Thus $g(2m,2n) \neq 2 g(m,n)$ in general. (Statement C is FALSE)

• Option (D): $g(2m,2n) = 2^{2(m+n)} = (2^{m+n})^2 = (g(m,n))^2$ Hence, $g(2m,2n) = (g(m,n))^2$ for all positive integers $m, n$. (Statement D is TRUE)

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Final Conclusion:

The correct statements are A, B, and D.