For non-negative integers s and r, let s r = ll s! r!(s-r)! & if …

MCQ

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 $m, n$

✓

Answer

Correct option: A.

$A,B,D$

a
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

$g ( m , n )=\sum_{ p =0}^{ m + n } \frac{f( m , n , p )}{{ }^{ n + p } C _{ p }}$

$g ( m , n )=\sum_{ p =0}^{ m + n + m + n } C _{ p }$

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

$(A)$ $g(m, n)=q(n, m)$

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

$g ( m + n , n )=2^{ m +1+ n }$

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

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

$=( g ( m , n ))^2$

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