For nonnegative integers s and r , let ( array ll s r array ) = \...

f\left(m, n, p\right) = \mathop{∑}_{i = 0}^{p}\left($$\begin{array}{ll}m \\ i\end{array}$$\right)\left($$\begin{array}{ll}n + i \\ p\end{array}$$\right)\left($$\begin{array}{ll}p + n \\ p - i\end{array}$$\right)$</p><p>$= \mathop{∑}_{i = 0}^{p} Ci m \cdot \dfrac{\left(n + i\right) !}{p ! \left(n + i - p\right) !} \cdot \dfrac{( p + n ) !}{\left(p - i\right) ! \left(n + i\right) !}$</p><p>$= \mathop{∑}_{i = 0}^{p} Ci m \cdot \dfrac{( p + n ) !}{p ! n !} \cdot \dfrac{n !}{\left(n + i - p\right) ! \left(p - i\right) !}$</p><p>$= \mathop{∑}_{i = 0}^{p} Ci m . Cp p + n \cdot Cp - i n$</p><p>$Cp p + n \cdot \mathop{∑}_{i = 0}^{p} Ci m \cdot Cp - i n$</p><p>\left(∵ \mathop{∑}_{i = 0}^{p} Ci m \cdot Cp - i n = Cp m + n\right)$

$∴ f\left(m, n, p\right) = Cp p + n . Cp m + n$

$⇒ \dfrac{f\left(m, n, p\right)}{Cp p + n} = Cp m + n$.

$g\left(m, n\right) = \mathop{∑}_{p = 0}^{m + n} Cp m + n = {2}^{m + n}$

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

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

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

(C) $g\left(2m, 2n\right) = {2}^{2m + 2n} = {\left({2}^{m + n}\right)}^{2} = {\left(g\left(m, n\right)\right)}^{2}$

(D) $g\left(m, n\right) = g\left(n, m\right)$.