The number of ways of dividing 15 persons int | vedclass

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(D) Total persons = $15$. Groups required are of sizes $3, 5, 7$. Let the two particular persons be $P_1$ and $P_2$. We need to form groups such that

Vedclass · Accepted Answer

$13 	imes \frac{11!}{3!5!}$

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(D) Total persons = $15$. Groups required are of sizes $3, 5, 7$. Let the two particular persons be $P_1$ and $P_2$. We need to form groups such that $P_1$ and $P_2$ are not in the group of $5$. Case $1$: Both $P_1$ and $P_2$ are in the group of $3$. Remaining $13$ persons are to be divided into groups of $1, 5, 7$. Number of ways = $\binom{13}{1} 	imes \binom{12}{5} 	imes \binom{7}{7} = 13 	imes \frac{12!}{5!7!} = 13 	imes \frac{12 	imes 11!}{5!7!} = \frac{156 	imes 11!}{5!7!}$. Case $2$: Both $P_1$ and $P_2$ are in the group of $7$. Remaining $13$ persons are to be divided into groups of $3, 5, 5$. Number of ways = $\binom{13}{3} 	imes \binom{10}{5} 	imes \binom{5}{5} = \frac{13!}{3!10!} 	imes \frac{10!}{5!5!} = \frac{13!}{3!5!5!}$. Case $3$: One is in the group of $3$ and one is in the group of $7$. Number of ways = $\binom{2}{1} 	imes \binom{13}{2} 	imes \binom{11}{4} 	imes \binom{7}{6} = 2 	imes \frac{13!}{2!11!} 	imes \frac{11!}{4!7!} 	imes 7 = \frac{13!}{4!6!}$. Summing these cases or using the complement method: Total ways to divide $15$ into $3, 5, 7$ is $\frac{15!}{3!5!7!}$. Ways where $P_1, P_2$ are in the group of $5$ is $\binom{13}{3} 	imes \binom{10}{7} = \frac{13!}{3!10!} 	imes \frac{10!}{7!3!} = \frac{13!}{3!3!7!}$. Subtracting this from total: $\frac{15!}{3!5!7!} - \frac{13!}{3!3!7!} = \frac{13!}{3!7!} [\frac{15 	imes 14}{5 	imes 4} - \frac{1}{3}] = \frac{13!}{3!7!} [4.2 - 0.33] = 13 	imes \frac{11!}{3!5!}$.

The number of ways of dividing $15$ persons into $3$ groups containing $3, 5$ and $7$ persons such that two particular persons are not included in the $5$-person group is:

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