JEE Advanced 2024 Paper 1 · Q11 · Permutations & Combinations

Question

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 ______.

SolveKar AI · Accepted Answer

Use complementary counting. Total ways without any restriction: $\binom{9}{2}\binom{7}{3}\binom{4}{4}=36\cdot 35\cdot 1=1260$. Inclusion-exclusion. Let $A$: $s_1\in X$; $B$: $s_2\in Y$. $|A|$: fix $s_1$ in $X$, choose remaining: $\binom{8}{1}\binom{7}{3}\binom{4}{4}=8\cdot 35=280$. $|B|$: fix $s_2$ in $Y$: $\binom{8}{2}\binom{6}{2}\binom{4}{4}=28\cdot 15=420$. $|A\cap B|$: fix $s_1\in X$, $s_2\in Y$: $\binom{7}{1}\binom{6}{2}\binom{4}{4}=7\cdot 15=105$. Valid $=1260-280-420+105=665$.

JEE Advanced 2024 Paper 1 · Q11 · Permutations & Combinations

Solution