For non-negative integers s and r,let binom{s | vedclass

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Question

(A) Given $f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$.Using the identity $\binom{n+i}{p} \binom{p+n}{p-i} = \binom{n+p}

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(A) Given $f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$.Using the identity $\binom{n+i}{p} \binom{p+n}{p-i} = \binom{n+p}{p} \binom{n+i}{i}$,we have:$f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+p}{p} \binom{n+i}{i} = \binom{n+p}{p} \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{i}$.Using Vandermonde's Identity or combinatorial simplification,$\sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{i} = \binom{m+n+1}{p}$ is not directly applicable,but rather $\sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p-i}$ simplifies to $\binom{m+n+p}{p}$.Actually,$f(m, n, p) = \binom{n+p}{p} \binom{m+n}{p}$.Thus,$\frac{f(m, n, p)}{\binom{n+p}{p}} = \binom{m+n}{p}$.Then $g(m, n) = \sum_{p=0}^{m+n} \binom{m+n}{p} = 2^{m+n}$.Checking the options:$(A)$ $g(m, n) = 2^{m+n}$ and $g(n, m) = 2^{n+m}$,so $g(m, n) = g(n, m)$ is $TRUE$.$(B)$ $g(m, n+1) = 2^{m+n+1}$ and $g(m+1, n) = 2^{m+1+n}$,so $g(m, n+1) = g(m+1, n)$ is $TRUE$.$(C)$ $g(2m, 2n) = 2^{2m+2n}$ and $2g(m, n) = 2 \cdot 2^{m+n} = 2^{m+n+1}$,so $g(2m, 2n) 
eq 2g(m, n)$.$(D)$ $g(2m, 2n) = 2^{2m+2n} = (2^{m+n})^2 = (g(m, n))^2$,so $g(2m, 2n) = (g(m, n))^2$ is $TRUE$.Therefore,the correct statements are $A, B, D$.

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