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Mathematics · Binomial Theorem

JEE Advanced 2020 — Paper 2 — Question 45

For nonnegative integers s and r, let (sr)={s!r!(s−r)!if 0≤r≤s,0if r>s.\binom{s}{r} = \begin{cases} \frac{s!}{r!(s-r)!} & \text{if } 0 \leq r \leq s, \\ 0 & \text{if } r > s. \end{cases} For positive integers m and n, let g(m,n)=∑p=0m+nf(m,n,p)(n+pp)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)=∑i=0p(mi)(n+ip)(p+np−i)f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}

  1. Option A:

    g(m,n)=g(n,m)g(m, n)=g(n, m) for all positive integers m,nm, n

    Correct
  2. Option B:

    g(m,n+1)=g(m+1,n)g(m, n+1)=g(m+1, n) for all positive integers m,nm, n

    Correct
  3. Option C:

    g(2m,2n)=2g(m,n)g(2 m, 2 n)=2 g(m, n) for all positive integers m,nm, n

  4. Option D:

    g(2m,2n)=(g(m,n))2g(2 m, 2 n)=(g(m, n))^{2} for all positive integers m,nm, n

Answer: A, B

Step-by-step solution

f(m,n,p)=∑i=0p(mCin+iCpp+nCp−i)=∑i=0p(m!i!(m−1)!⋅(n+i)!p!(n+i−p)!⋅(p+n)!(p−i)!(n+i)!)=∑i=0pmCi⋅n+pCn⋅nCp−i=n+pCn.m+nCpg(m,n)=∑p=0m+nmm+nCp.n+pCnn+pCp∑p=0m+nm+nCpg(m,n)=2m+n. \begin{aligned} & f(m, n, p)=\sum_{i=0}^{p}\left({ }^{m} C_{i}{ }^{n+i} C_{p}{ }^{p+n} C_{p-i}\right) \\& =\sum_{i=0}^{p}\left(\frac{m!}{i!(m-1)!} \cdot \frac{(n+i)!}{p!(n+i-p)!} \cdot \frac{(p+n)!}{(p-i)!(n+i)!}\right)=\sum_{i=0}^{p}{ }^{m} C_{i} \cdot{ }^{n+p} C_{n} \cdot{ }^{n} C_{p-i} \\& ={ }^{n+p} C_{n} .{ }^{m+n} C_{p} \\& g\left(m, n\right)=\sum_{p=0}^{m+n} \frac{m^{m+n} C_{p} .{ }^{n+p} C_{n}}{n+p} C_{p} \sum_{p=0}^{m+n}{ }^{m+n} C_{p} \\& g(m,n)=2^{\mathrm{m}+\mathrm{n}} \text {. } \end{aligned}

A,B and D satisfied.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Mathematics
Chapter
Binomial Theorem
Topic
Binomial Coefficients