Question

For nonnegative integers $s$ and $r,$ let $\left(\begin{array}{c}s\\ r\end{array}\right)=\{\begin{array}{cc}\frac{s!}{r!(s-r)!}& \text{if}r\le s,\\ 0& \text{if}r>s.\end{array}$

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?

• A. $g(m,n)=g(n,m)$ for all positive integers $m,n$
• B. $g(m,n+1)=g(m+1,n)$ for all positive integers $m,n$
• C. $g(2m,2n)=2g(m,n)$ for all positive integers $m,n$
• D. $g(2m,2n)={\left(g(m,n)\right)}^2$ for all positive integers $m,n$

Answer 1

Answer: (A), (B), (D)

$g(2m,2n)={\left(g(m,n)\right)}^2$ for all positive integers $m,n$

$f(m,n,p)=\sum _{i=0}^p\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}n+i\\ p\end{array}\right)\left(\begin{array}{c}p+n\\ p-i\end{array}\right)$
$=\sum _{i=0}^p{}^m{C}_i\cdot {}^{n+i}{C}_p\cdot {}^{n+p}{C}_{p-i}[\because \left(\begin{array}{c}n\\ r\end{array}\right)={}^n{C}_r]$
$=\sum _{i=0}^p{}^m{C}_i\frac{(n+i)!}{\left(p\right)!(n+i-p)!}\frac{(n+p)!}{(p-i)!(n+p-p+i)!}$
$=\sum _{i=0}^p{}^m{C}_i\left(\frac{(n+i)!}{\left(p\right)!(n+i-p)!}\right)\left(\frac{(n+p)!}{(p-i)!(n+i)!}\right)$
$=\sum _{i=0}^p{}^m{C}_i\left(\frac{(n+p)!}{\left(p\right)!(n+i-p)!(p-i)!}\right)$
$=\sum _{i=0}^p{}^m{C}_i\left(\frac{(n+p)!}{\left(p\right)!\left(n\right)!}\right)\left(\frac{\left(n\right)!}{(n+i-p)!(p-i)!}\right)$
$={}^{n+p}{C}_p\left[\sum _{i=0}^p{}^m{C}_i\cdot {}^n{C}_{p-i}\right]$
$={}^{n+p}{C}_p[{}^m{C}_0\cdot {}^n{C}_p+{}^m{C}_1\cdot {}^n{C}_{p-1}\cdots {}^m{C}_p\cdot {}^n{C}_0]$

$\therefore f(m,n,p)=\left({}^{n+p}{C}_p\right)\left({}^{m+n}{C}_p\right)$

$g(m,n)=\sum _{p=0}^{m+n}\frac{f(m,n,p)}{\left(\begin{array}{c}n+p\\ p\end{array}\right)}=\sum _{p=0}^{m+n}\frac{\left({}^{n+p}{C}_p\right)\left({}^{m+n}{C}_p\right)}{\left({}^{n+p}{C}_p\right)}$
$\Rightarrow g(m,n)=\sum _{p=0}^{m+n}{}^{m+n}{C}_p=2^{m+n}=2^{n+m}$
$g(m,n)=g(n,m)$
$g(2m,2n)=2^{2(m+n)}={\left(2^{m+n}\right)}^2={\left(g(m,n)\right)}^2$
$g(m,n+1)=2^{m+n+1}=2^{(m+1)+n}=g(m+1,n)$