JEE Advanced 2020 · Paper 2 · Mathematics Q12

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MathematicsMultiple correct+4 / -2Question 48 of 54

For nonnegative integers $s$ and $r$ , let

$\binom sr = \begin{cases}\dfrac{s!}{r!\,(s-r)!} & \text{if } r \le s,\\ 0 & \text{if } r > s.\end{cases}$

For positive integers $m$ and $n$ , let

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

where for any nonnegative integer $p$ ,

$f(m,n,p) = \sum_{i=0}^{p}\binom mi\binom{n+i}{p}\binom{p+n}{p-i}.$

Then which of the following statements is/are TRUE?

g(m,n) = g(n,m)

for all positive integers

m, n

g(m,n+1) = g(m+1,n)

for all positive integers

m, n

g(2m,2n) = 2\,g(m,n)

for all positive integers

m, n

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

for all positive integers

m, n

Answer and worked solution

Official answer: A, B, D

Simplify the inner product. For $i\le p$ :
$\binom{n+i}p\binom{p+n}{p-i}=\frac{(n+i)!}{p!\,(n+i-p)!}\cdot\frac{(p+n)!}{(p-i)!\,(n+i)!}=\binom{n+p}p\binom n{p-i}.$

Apply Vandermonde.
$\frac{f(m,n,p)}{\binom{n+p}p}=\sum_i\binom mi\binom n{p-i}=\binom{m+n}p.$
So
$g(m,n)=\sum_{p=0}^{m+n}\binom{m+n}p=2^{m+n}.$

The statements.

(A) $2^{m+n}$ is symmetric. True.
(B) Both sides are $2^{m+n+1}$ . True.
(C) $g(2m,2n)=2^{2(m+n)}\ne2\cdot2^{m+n}$ . False.
(D) $2^{2(m+n)}=(2^{m+n})^2$ . True.

(A), (B), (D)

Topics:Properties of binomial coefficients Combinations

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JEE Advanced 2020 · Paper 2, Mathematics question 12