JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Permutations and Combinations

Total students: 9. Teams: X (2), Y (3), Z (4). Constraints: $s_1\notin X$, $s_2\notin Y$.

**Case I: $s_2\in X$**

X contains $s_2$ and one from the remaining 7 (excluding $s_1$): $\left(\genfrac{}{}{0ex}{}{7}{1}\right)=7$ ways.

Remaining 7 students (including $s_1$) are divided into Y (3) and Z (4): $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ ways.

Total: $7\times 35=245$ ways.

**Case II: $s_2\notin X$**

X is chosen from 7 students (excluding $s_1,s_2$): $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ ways.

Remaining 7 students include $s_1,s_2$. Y must not contain $s_2$, so Y is chosen from 6 students (excluding $s_2$): $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ ways.

Total: $21\times 20=420$ ways.

**Grand total: $245+420=665$ ways.**