JEE Main 2026 — 2 April, Evening Shift — Question 28

We need to evaluate $S = \binom{3}{30-r} + 3\binom{3}{31-r} + 3\binom{3}{32-r} + \binom{3}{33-r}$. Note that $\binom{3}{k}$ is zero unless $0 \le k \le 3$.

Since $3 \le r \le 30$, we have $30-r \le 27$,

so only terms with $30-r, 31-r, 32-r, 33-r$ in $\{0,1,2,3\}$ are non-zero.

But we can use the identity $\binom{3}{0}\binom{30}{30-r} + \binom{3}{1}\binom{30}{31-r} + \binom{3}{2}\binom{30}{32-r} + \binom{3}{3}\binom{30}{33-r} = \binom{33}{33-r}$ by Vandermonde's identity. Thus $S = \binom{33}{33-r}$. Using symmetry $\binom{33}{33-r} = \binom{33}{r}$. Hence $\binom{33}{r} = \binom{m}{r}$, so $m = 33$.