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

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Binomial Theorem

The number of elements in the set

S = {(r, k) : k \in Z

and

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

, is

Answer: (B)

Using

^{36} C_{r + 1} = \frac{36}{r + 1}^{35} C_{r}

and cancelling

^{35} C_{r}

, the relation becomes

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

, so

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

, i.e.

k^{2} = \frac{r + 19}{6}

. For

0 \leq r \leq 35

this gives

\frac{19}{6} \leq k^{2} \leq \frac{54}{6}

, that is

3.17 \leq k^{2} \leq 9

. The only perfect squares in this range are

k^{2} = 4

(giving

r = 5

) and

k^{2} = 9

(giving

r = 35

). Each admits

k

of either sign, producing the pairs

(5, 2), (5, - 2), (35, 3), (35, - 3)

, so the set has

4

elements.

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option