JEE Advanced 2024 — Paper 1 — MATH-III solution

JEE Advanced2024Paper 1MATH-III

A group of 9 students, $s_{1}, s_{2}, \dots, s_{9}$ , is to be divided to form three teams $X$ , $Y$ and $Z$ of sizes 2, 3, and 4, respectively. Suppose that $s_{1}$ cannot be selected for the team $X$ , and $s_{2}$ cannot be selected for the team $Y$ . Then the number of ways to form such teams, is ______.

Your answer

Solution

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 $= 280 + 420 - 105 = 595$ .

$∴ Required = 1260 - 595 = 665$

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