JEE Main 2026 Apr 2 Shift 1, Mathematics Q7: Combinations

JEE Main2026Apr 2, Shift 1Mathematics

The number of elements in the set $S = {(r, k) : k \in Z and^{36} C_{r + 1} = \frac{6 \cdot ^{35} C _{r}}{k ^{2} - 3}}$ , is

Solution

Using the identity $^{36} C_{r + 1} = \frac{36}{r + 1} \cdot^{35} C_{r}$ , the given equation becomes:

$\frac{36}{r + 1} = \frac{6}{k ^{2} - 3} \Rightarrow k^{2} - 3 = \frac{r + 1}{6} \Rightarrow k^{2} = \frac{r + 19}{6}$

With $0 \leq r \leq 35$ (so that $^{35} C_{r}$ is defined), $19 \leq r + 19 \leq 54$ , hence $k^{2} \in [19/6, 9] = [3.16, 9]$ .

The only integer perfect squares in this range are $4$ and $9$ , giving $k = \pm 2$ (then $r = 5$ ) and $k = \pm 3$ (then $r = 35$ ).

The valid pairs are $(5, 2), (5, - 2), (35, 3), (35, - 3)$ — total $4$ elements.

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.