The user wants me to solve a mathematics problem involving summations of binomial coefficients and determine which of the given statements are true.

Let's first analyze and simplify the given expressions for f(m, n, p) and g(m, n).

Step 1: Simplify f(m, n, p)

The function f(m, n, p) is defined as: $f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\begin{matrix} m \\ i \\ \end{matrix} } \right)\left( {\begin{matrix} {n + i} \\ p \\ \end{matrix} } \right)\left( {\begin{matrix} {p + n} \\ {p - i} \\ \end{matrix} } \right)}$

Let's analyze the product of the last two binomial coefficients in the summand. We can use the identity $\left( {\begin{matrix} n \\ k \\ \end{matrix} } \right) = \left( {\begin{matrix} n \\ {n - k} \\ \end{matrix} } \right)$.

$\left( {\begin{matrix} {p + n} \\ {p - i} \\ \end{matrix} } \right) = \left( {\begin{matrix} {p + n} \\ {(p+n) - (p - i)} \\ \end{matrix} } \right) = \left( {\begin{matrix} {p + n} \\ {n + i} \\ \end{matrix} } \right)$

So the product becomes: $\left( {\begin{matrix} {n + i} \\ p \\ \end{matrix} } \right)\left( {\begin{matrix} {p + n} \\ {n + i} \\ \end{matrix} } \right)$

Now, we use the identity $\left( {\begin{matrix} a \\ b \\ \end{matrix} } \right)\left( {\begin{matrix} b \\ c \\ \end{matrix} } \right) = \left( {\begin{matrix} a \\ c \\ \end{matrix} } \right)\left( {\begin{matrix} {a - c} \\ {b - c} \\ \end{matrix} } \right)$. Let a = p + n, b = n + i, and c = p.

$\left( {\begin{matrix} {p + n} \\ {n + i} \\ \end{matrix} } \right) \left( {\begin{matrix} {n + i} \\ p \\ \end{matrix} } \right) = \left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right) \left( {\begin{matrix} {(p+n)-p} \\ {(n+i)-p} \\ \end{matrix} } \right) = \left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right) \left( {\begin{matrix} n \\ {n+i-p} \\ \end{matrix} } \right)$

Substitute this back into the expression for f(m, n, p):

$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\begin{matrix} m \\ i \\ \end{matrix} } \right) \left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right) \left( {\begin{matrix} n \\ {n+i-p} \\ \end{matrix} } \right)}$

The term $\left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right)$ is independent of i, so we can pull it out of the summation:

$f(m,n,p) = \left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right) \sum\limits_{i = 0}^p {\left( {\begin{matrix} m \\ i \\ \end{matrix} } \right) \left( {\begin{matrix} n \\ {n+i-p} \\ \end{matrix} } \right)}$

Using the identity $\left( {\begin{matrix} n \\ k \\ \end{matrix} } \right) = \left( {\begin{matrix} n \\ {n - k} \\ \end{matrix} } \right)$ on the second term in the sum:

$\left( {\begin{matrix} n \\ {n+i-p} \\ \end{matrix} } \right) = \left( {\begin{matrix} n \\ {n - (n+i-p)} \\ \end{matrix} } \right) = \left( {\begin{matrix} n \\ {p-i} \\ \end{matrix} } \right)$

The summation becomes: $\sum\limits_{i = 0}^p {\left( {\begin{matrix} m \\ i \\ \end{matrix} } \right) \left( {\begin{matrix} n \\ {p-i} \\ \end{matrix} } \right)}$

This is Vandermonde's Identity, which states $\sum\limits_{k=0}^r {\left( {\begin{matrix} m \\ k \\ \end{matrix} } \right) \left( {\begin{matrix} n \\ {r-k} \\ \end{matrix} } \right)} = \left( {\begin{matrix} {m+n} \\ r \\ \end{matrix} } \right)$. Here, r = p.

So, the sum is equal to $\left( {\begin{matrix} {m+n} \\ p \\ \end{matrix} } \right)$.

Therefore, the simplified expression for f(m, n, p) is: $f(m,n,p) = \left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right) \left( {\begin{matrix} {m+n} \\ p \\ \end{matrix} } \right)$

Step 2: Simplify g(m, n)

The function g(m, n) is defined as: $g(m,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\begin{matrix} {n + p} \\ p \\ \end{matrix} } \right)}}}$

Substitute the simplified f(m, n, p) into this expression:

$g(m,n) = \sum\limits_{p = 0}^{m + n} {{{\left( {\begin{matrix} {p + n} \\ p \\ \end{matrix} } \right)\left( {\begin{matrix} {m+n} \\ p \\ \end{matrix} } \right)} \over {\left( {\begin{matrix} {n + p} \\ p \\ \end{matrix} } \right)}}}$

The term $\left( {\begin{matrix} {n + p} \\ p \\ \end{matrix} } \right)$ cancels out from the numerator and denominator.

$g(m,n) = \sum\limits_{p = 0}^{m + n} {\left( {\begin{matrix} {m+n} \\ p \\ \end{matrix} } \right)}$

This is the sum of all binomial coefficients for m+n$. From the binomial theorem, we know that $\sum\limits_{k=0}^N {\left( {\begin{matrix} N \\ k \\ \end{matrix} } \right)} = 2^N$. Here, $N = m+n.

Thus, we have the very simple result: $g(m,n) = 2^{m+n}$

Step 3: Evaluate the options

Now we check each statement using $g(m, n) = 2^{m+n}$.

A: g(m, n) = g(n, m) for all positive integers m, n

• $g(m, n) = 2^{m+n}$
• $g(n, m) = 2^{n+m}$
• Since addition is commutative, m+n = n+m. Therefore, g(m, n) = g(n, m).
• Statement A is TRUE.

B: g(m, n + 1) = g(m + 1, n) for all positive integers m, n

• $g(m, n + 1) = 2^{m + (n+1)} = 2^{m+n+1}$
• $g(m + 1, n) = 2^{(m+1) + n} = 2^{m+1+n}$
• The exponents are equal. Therefore, g(m, n + 1) = g(m + 1, n).
• Statement B is TRUE.

C: g(2m, 2n) = 2g(m, n) for all positive integers m, n

• $g(2m, 2n) = 2^{2m + 2n} = 2^{2(m+n)} = (2^2)^{m+n} = 4^{m+n}$
• $2g(m, n) = 2 \cdot 2^{m+n} = 2^{m+n+1}$
• The equation $4^{m+n} = 2^{m+n+1}$ implies $2^{2(m+n)} = 2^{m+n+1}$. Equating the exponents gives 2(m+n) = m+n+1, which simplifies to m+n=1. Since m and n are positive integers, $m \ge 1$ and $n \ge 1$, so $m+n \ge 2$. Thus, the equality does not hold.
• Statement C is FALSE.

D: g(2m, 2n) = (g(m, n))^2 for all positive integers m, n

• $g(2m, 2n) = 2^{2m + 2n} = 2^{2(m+n)}$
• $(g(m, n))^2 = (2^{m+n})^2 = 2^{2(m+n)}$
• The expressions are equal. Therefore, $g(2m, 2n) = (g(m, n))^2$.
• Statement D is TRUE.

Final conclusion: Statements A, B, and D are true.