JEE Advanced 2024 · Paper 1 · Mathematics Q11

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MathematicsInteger+4Question 11 of 51

A group of 9 students, $s_1, s_2, \ldots, s_9$ , 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

Answer and worked solution

Official answer: 665

Without restrictions: $\dfrac{9!}{2!\,3!\,4!} = 1260$ .

Bad cases (by inclusion-exclusion):

$s_1 \in X$ : partner for $X$ from 8, then $Y$ from 7: $8\cdot\binom73 = 280$ .
$s_2 \in Y$ : two more for $Y$ from 8, then $X$ from 6: $\binom82\binom62 = 420$ .
Both: partner of $s_1$ in $X$ from the 7 others (not $s_2$ ), two more for $Y$ from 6: $7\cdot\binom62 = 105$ .

Valid ways $= 1260 - 280 - 420 + 105 = 665$ .

Topics:Distribution and arrangements Combinations

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JEE Advanced 2024 · Paper 1, Mathematics question 11