The set S = {1, 2, 3, dots, 12} is partitione | vedclass

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(B) The set $S$ contains $12$ elements. Since it is partitioned into three sets $A, B, C$ of equal size,each set must contain $12 / 3 = 4$ elements.Th

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

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(B) The set $S$ contains $12$ elements. Since it is partitioned into three sets $A, B, C$ of equal size,each set must contain $12 / 3 = 4$ elements.The number of ways to choose $4$ elements for set $A$ is $\binom{12}{4}$.The number of ways to choose $4$ elements for set $B$ from the remaining $8$ is $\binom{8}{4}$.The number of ways to choose $4$ elements for set $C$ from the remaining $4$ is $\binom{4}{4}$.Since the sets $A, B, C$ are distinct (labeled),the total number of ways is:$\binom{12}{4} 	imes \binom{8}{4} 	imes \binom{4}{4} = \frac{12!}{4! 	imes 8!} 	imes \frac{8!}{4! 	imes 4!} 	imes \frac{4!}{4! 	imes 0!} = \frac{12!}{(4!)^3}$.

The set $S = \{1, 2, 3, \dots, 12\}$ is partitioned into three sets $A, B, C$ of equal size such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = C \cap A = \phi$. In how many ways can $S$ be partitioned?

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