JEE Main 2026 — Binomial Theorem Question with Solution

Given equation:

${}^{36}{C}_{r+1}=\frac{6\left({}^{35}{C}_r\right)}{k^2-3}$

Using the property ${}^n{C}_r=\frac{n}{r}{}^{n-1}{C}_{r-1}$, we can write:

${}^{36}{C}_{r+1}=\frac{36}{r+1}{}^{35}{C}_r$

Substituting this into the given equation:

$\frac{36}{r+1}{}^{35}{C}_r=\frac{6}{k^2-3}{}^{35}{C}_r$

For ${}^{35}{C}_r$ to be defined, $r$ must be an integer such that $0\le r\le 35$. Thus, ${}^{35}{C}_r\ne 0$. Dividing both sides by ${}^{35}{C}_r$:

$\frac{36}{r+1}=\frac{6}{k^2-3}$

$\frac{6}{r+1}=\frac{1}{k^2-3}$

$k^2-3=\frac{r+1}{6}$

$k^2=3+\frac{r+1}{6}$

For $k\in \mathbb{Z}$, $k^2$ must be a perfect square integer. This requires $r+1$ to be a multiple of $6$.

Since $0\le r\le 35$, we have $1\le r+1\le 36$.

Let $r+1=6m$, where $m\in \{1,2,3,4,5,6\}$.

Then $k^2=3+m$.

Checking the possible values of $m$ for which $k^2$ is a perfect square:

For $m=1$, $k^2=4\Rightarrow k=\pm 2$. (Here $r+1=6\Rightarrow r=5$)

For $m=2$, $k^2=5$ (not a perfect square).

For $m=3$, $k^2=6$ (not a perfect square).

For $m=4$, $k^2=7$ (not a perfect square).

For $m=5$, $k^2=8$ (not a perfect square).

For $m=6$, $k^2=9\Rightarrow k=\pm 3$. (Here $r+1=36\Rightarrow r=35$)

The possible pairs $(r,k)$ are $(5,2)$, $(5,-2)$, $(35,3)$, and $(35,-3)$.

Therefore, the number of elements in the set $S$ is $4$.

Answer: $4$