Binomial TheoremmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Binomial Identity Set with Integer k: 4 Elements | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The number of elements in the set S={(r,k):kZ and 36Cr+1=6(35Cr)k23}S=\left\{(r,k):k\in\mathbb{Z}\ \text{and}\ {}^{36}C_{r+1}=\dfrac{6\left({}^{35}C_r\right)}{k^2-3}\right\} is
A22
B44correct
C88
D1616
Solution
Step 1: Use 36Cr+1=36r+135Cr{}^{36}C_{r+1}=\dfrac{36}{r+1}\,{}^{35}C_r. Substitute:
36r+135Cr=635Crk23  36r+1=6k23.\frac{36}{r+1}\,{}^{35}C_r=\frac{6\,{}^{35}C_r}{k^2-3}\ \Rightarrow\ \frac{36}{r+1}=\frac{6}{k^2-3}.
Step 2: Cross-multiply:
k23=6(r+1)36=r+16  k2=r+196.k^2-3=\frac{6(r+1)}{36}=\frac{r+1}{6}\ \Rightarrow\ k^2=\frac{r+19}{6}.
Step 3: Constraint 0r350\le r\le 35 gives
0+196k235+196  3.16k29.\frac{0+19}{6}\le k^2\le\frac{35+19}{6}\ \Rightarrow\ 3.16\ldots\le k^2\le 9.
So the perfect-square integer values are k2=4k^2=4 and k2=9k^2=9. Step 4: k2=4r+19=24r=5k^2=4\Rightarrow r+19=24\Rightarrow r=5, with k=±2k=\pm2. k2=9r+19=54r=35k^2=9\Rightarrow r+19=54\Rightarrow r=35, with k=±3k=\pm3. Step 5: The valid pairs are (5,2),(5,2),(35,3),(35,3)(5,2),(5,-2),(35,3),(35,-3) — that is 44 elements. Correct answer: (2)
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