First slide

Combinations

Question

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

Moderate

A

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

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

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

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

Solution

Each of the three sets $A, B, C$ contains exactly $4$ elements.
Thus, the number of ways of partitioning the set $S$ is
$\frac{1}{3!} (^{12} C_{4}) (^{8} C_{4}) (^{4} C_{4}) = \frac{1}{3!} \frac{12!}{4! 8!} \times \frac{8!}{4! 4!} (1) = \frac{12!}{3! (4!)^{3}}$

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