JEE Main 2026 — Mathematics Algebra

Given equation:

$$^{36}C_{r+1} = \dfrac{6 \left(^{35}C_r\right)}{k^2 - 3}$$36Cr+1​=k2−36(35Cr​)​

Using the property $^{n}C_r = \dfrac{n}{r} {}^{n-1}C_{r-1}$nCr​=rn​n−1Cr−1​, we can write:

$$^{36}C_{r+1} = \dfrac{36}{r+1} {}^{35}C_r$$36Cr+1​=r+136​35Cr​

Substituting this into the given equation:

$$\dfrac{36}{r+1} {}^{35}C_r = \dfrac{6}{k^2 - 3} {}^{35}C_r$$r+136​35Cr​=k2−36​35Cr​

For $^{35}C_r$35Cr​ to be defined, $r$r must be an integer such that $0 \le r \le 35$0≤r≤35. Thus, $^{35}C_r \neq 0$35Cr​=0. Dividing both sides by $^{35}C_r$35Cr​:

$$\dfrac{36}{r+1} = \dfrac{6}{k^2 - 3}$$r+136​=k2−36​

$$\dfrac{6}{r+1} = \dfrac{1}{k^2 - 3}$$r+16​=k2−31​

$$k^2 - 3 = \dfrac{r+1}{6}$$k2−3=6r+1​

$$k^2 = 3 + \dfrac{r+1}{6}$$k2=3+6r+1​

For $k \in \mathbb{Z}$k∈Z, $k^2$k2 must be a perfect square integer. This requires $r+1$r+1 to be a multiple of $6$6.

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

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

Then $k^2 = 3 + m$k2=3+m.

Checking the possible values of $m$m for which $k^2$k2 is a perfect square:

For $m = 1$m=1, $k^2 = 4 \Rightarrow k = \pm 2$k2=4⇒k=±2. (Here $r+1 = 6 \Rightarrow r = 5$r+1=6⇒r=5)

For $m = 2$m=2, $k^2 = 5$k2=5 (not a perfect square).

For $m = 3$m=3, $k^2 = 6$k2=6 (not a perfect square).

For $m = 4$m=4, $k^2 = 7$k2=7 (not a perfect square).

For $m = 5$m=5, $k^2 = 8$k2=8 (not a perfect square).

For $m = 6$m=6, $k^2 = 9 \Rightarrow k = \pm 3$k2=9⇒k=±3. (Here $r+1 = 36 \Rightarrow r = 35$r+1=36⇒r=35)

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

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

Answer: $4$4