The set S = {1, 2, 3, dots, 12} is to be part | vedclass

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(C) The set $S$ contains $12$ elements.We need to partition $S$ into three disjoint sets $A, B, C$ of equal size.Since $|S| = 12$,each set must contai

Vedclass · Accepted Answer

$\frac{12!}{3!(4!)^3}$

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(C) The set $S$ contains $12$ elements.We need to partition $S$ into three disjoint sets $A, B, C$ of equal size.Since $|S| = 12$,each set must contain $12 / 3 = 4$ elements.The number of ways to distribute $12$ distinct items into $3$ unlabeled groups of size $4$ is given by the formula $\frac{1}{3!} \binom{12}{4, 4, 4}$.This is calculated as:$\frac{1}{3!} 	imes \binom{12}{4} 	imes \binom{8}{4} 	imes \binom{4}{4} = \frac{1}{3!} 	imes \frac{12!}{4!8!} 	imes \frac{8!}{4!4!} 	imes \frac{4!}{4!0!} = \frac{12!}{3!(4!)^3}$.

The set $S = \{1, 2, 3, \dots, 12\}$ is to be partitioned into three sets $A, B, C$ of equal size such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = C \cap A = \emptyset$. The number of ways to partition $S$ is:

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