The set S = {1, 2, 3, ldots, 12} is to be par | 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 the total number of elements is

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

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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 the total number of elements is $12$,each set must contain $12 / 3 = 4$ elements.The number of ways to choose $4$ elements for set $A$ from $12$ 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 $\binom{4}{4}$.The total number of ways to distribute the elements into three labeled sets $A, B, C$ is $\binom{12}{4} 	imes \binom{8}{4} 	imes \binom{4}{4} = \frac{12!}{4!8!} 	imes \frac{8!}{4!4!} 	imes \frac{4!}{4!0!} = \frac{12!}{(4!)^3}$.Since the sets $A, B, C$ are not labeled (the partition is unordered),we must divide by $3!$ to account for the permutations of the three sets.Therefore,the total number of ways to partition $S$ is $\frac{12!}{3!(4!)^3}$.

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

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