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For non-negative integers $s$ and $r$, let
$\binom{s}{r}=\left\{\begin{array}{ll}\frac{s!}{r!(s-r)!} & \text { if } r \leq s \\ 0 & \text { if } r>s\end{array}\right.$
For positive integers $m$ and $n$, let
$(m, n) \sum_{ p =0}^{ m + n } \frac{ f ( m , n , p )}{\binom{ n + p }{ p }}$
where for any nonnegative integer $p$,
$f(m, n, p)=\sum_{i=0}^{ p }\binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}$
Then which of the following statements is/are $TRUE$?
$(A)$ $(m, n)=g(n, m)$ for all positive integers $m, n$
$(B)$ $(m, n+1)=g(m+1, n)$ for all positive integers $m, n$
$(C)$ $(2 m, 2 n)=2 g(m, n)$ for all positive integers $m, n$
$(D)$ $(2 m, 2 n)=(g(m, n))^2$ for all positive integers $m, n$
$A,B,D$
$A,B,C$
$A,B$
$A,D$
Solution
Solving
$f( m , n , p )=\sum_{ i =0}^{ p }{ }_{ m } C _{ i }^{ n + i } C _{ p } \cdot{ }^{ p + n } C _{ p – i }$
${ }^{ m } C _{ i } \cdot{ }^{ n + i } C _{ p \cdot}{ }^{ p + n } C _{ p – i }$
${ }^{ m } C _{ i } \cdot \frac{( n + i )!}{ p !( n – p + i )!} \times \frac{( n + p )!}{( p – i )!( n + i )!}$
${ }^{ m } C _1 \times \frac{( n + p )!}{ p !} \times \frac{1}{( n – p + i )!( p – i )!}$
${ }^{ m } C _{ i } \times \frac{( n + p )!}{ p ! n !} \times \frac{ n !}{( n – p + i )!( p – i )!}$
${ }^{ m } C _1{ }^{ n + p } C _{ p } \cdot{ }^{ n } C _{ p -1}\left\{{ }^{ m } C _{ e } \cdot{ }^{ n } C _{ p – i }={ }^{ m + n } C _{ p }\right\}$
$f( m , n , p )={ }^{ n + p } C _{ p } \cdot{ }^{ m + n } C _{ p }$
$\frac{f( m , n , p )}{{ }^{ n + p } C _{ p }}={ }^{ m + n } C _{ p }$
Now
$g ( m , n )=\sum_{ p =0}^{ m + n } \frac{f( m , n , p )}{{ }^{ n + p } C _{ p }}$
$g ( m , n )=\sum_{ p =0}^{ m + n + m + n } C _{ p }$
$g ( m , n )=2^{ m + n }$
$(A)$ $g(m, n)=q(n, m)$
$(B)$ $g(m, n+1)=2^{m+n+1}$
$g ( m + n , n )=2^{ m +1+ n }$
$(D)$ $g (2 m , 2 n )=2^{2 m +2 n }$
$=\left(2^{ m + n }\right)^2$
$=( g ( m , n ))^2$
