JEE Advanced 2024 Paper 1, Mathematics Section 3 Q4: Combinations

The total number of unrestricted ways to partition $9$ students into ordered groups of sizes $2,3,4$ is $\frac{9!}{2!3!4!}=1260$.

Use inclusion-exclusion on the forbidden events $A=\{s_1\in X\}$ and $B=\{s_2\in Y\}$.

$|A|$: fix $s_1$ in $X$, then arrange remaining 8 in sizes $1,3,4$: $\frac{8!}{1!3!4!}=280$.

$|B|$: fix $s_2$ in $Y$, then arrange remaining 8 in sizes $2,2,4$: $\frac{8!}{2!2!4!}=420$.

$|A\cap B|$: fix $s_1$ in $X$ and $s_2$ in $Y$, arrange remaining 7 in sizes $1,2,4$: $\frac{7!}{1!2!4!}=105$.

By inclusion-exclusion, forbidden count $=280+420-105=595$.

$$\therefore \text{Required}=1260-595=665$$