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 si

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6.Permutation and Combination

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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. . . .

A

$660$

B

$661$

C

$664$

D

$665$

(IIT-2024)

Solution

$x$ $y$ $z$

$2$ $3$ $4$

$\overline{ S }_1$ & $\overline{ S }_2$

$C$-i) when $x$ does not contain $S _1$, but contains $S _2$

${ }_{\text {for } x}^7 \times \frac{7!}{3!4!}=245$

for $x$ for $y . z$

C-ii) When $x$ does not contain $S _1, S _2$ and $y$ does not contain $S _2$ i.e. ${ }^7 C _2 \times \frac{6!}{3!3!}=420$

$\text { forx fory. } z$

so total No. of ways $665$

Standard 11

Mathematics

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