JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 7

Identity. ${}^{36}C_{r+1}=\dfrac{36}{r+1}\,{}^{35}C_r$, so the condition is $\dfrac{36}{r+1}=\dfrac6{k^2-3}$, i.e. $r+1=6(k^2-3)$.

Range. $1\le r+1\le36$ gives $1\le k^2-3\le6$, so $k^2\in\{4,9\}$:

• $k=\pm2$: $r=5$;
• $k=\pm3$: $r=35$.

4 elements.

(B)