A group of 9 students,s_1, s_2, ldots, s_9,is | vedclass

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(D) Total ways to form teams $X, Y, Z$ of sizes $2, 3, 4$ without any restrictions is $\binom{9}{2} 	imes \binom{7}{3} 	imes \binom{4}{4} = 36 	ime

Vedclass · Accepted Answer

$665$

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(D) Total ways to form teams $X, Y, Z$ of sizes $2, 3, 4$ without any restrictions is $\binom{9}{2} 	imes \binom{7}{3} 	imes \binom{4}{4} = 36 	imes 35 = 1260$.Let $A$ be the set of ways where $s_1 \in X$ and $B$ be the set of ways where $s_2 \in Y$.We want to find the total ways minus the ways where ($s_1 \in X$ $OR$ $s_2 \in Y$).By the Principle of Inclusion-Exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$.$|A|$ (ways where $s_1 \in X$): $s_1$ is fixed in $X$,we choose $1$ more from $8$ for $X$,then $3$ from $7$ for $Y$: $\binom{8}{1} 	imes \binom{7}{3} = 8 	imes 35 = 280$.$|B|$ (ways where $s_2 \in Y$): $s_2$ is fixed in $Y$,we choose $2$ from $8$ for $X$,then $2$ from $6$ for $Y$: $\binom{8}{2} 	imes \binom{6}{2} = 28 	imes 15 = 420$.$|A \cap B|$ (ways where $s_1 \in X$ $AND$ $s_2 \in Y$): $s_1$ fixed in $X$,$s_2$ fixed in $Y$,we choose $1$ from $7$ for $X$,then $2$ from $6$ for $Y$: $\binom{7}{1} 	imes \binom{6}{2} = 7 	imes 15 = 105$.$|A \cup B| = 280 + 420 - 105 = 595$.Total valid ways = $1260 - 595 = 665$.

$A$ group of $9$ students,$s_1, s_2, \ldots, s_9$,is to be divided into three teams $X, Y$,and $Z$ of sizes $2, 3$,and $4$,respectively. Suppose that $s_1$ cannot be selected for team $X$,and $s_2$ cannot be selected for team $Y$. The number of ways to form such teams is:

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