The set S={1,2,3−−−−−−−12} is to be portioned into three sets A, B, C of equal size. Thus A∪B∪C=S, A∩B=B∩C=A∩C=ϕ, the number of ways to partition S is

S{1,2−−−−−−12} A∪B∪C=S,A∩B=B∩C=C∩A=ϕ n(A)=n(B)=n(C)=4 (equal size)no. of ways =12C4.8C4.4C4=12!(4!)3

The set S={1,2,3−−−−−−−12} is to be portioned into three sets A, B, C of equal size. Thus A∪B∪C=S, A∩B=B∩C=A∩C=ϕ, the number of ways to partition S is

S{1,2−−−−−−12} A∪B∪C=S,A∩B=B∩C=C∩A=ϕ n(A)=n(B)=n(C)=4 (equal size)no. of ways =12C4.8C4.4C4=12!(4!)3

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

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

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

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

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

answer is C.

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Detailed Solution

$S {1, 2 - - - - - - 12}$
$A \cup B \cup C = S, A \cap B = B \cap C = C \cap A = ϕ$
$n (A) = n (B) = n (C) = 4$ (equal size)
no. of ways $=^{12} C_{4} .^{8} C_{4} .^{4} C_{4} = \frac{12!}{{(4!)}^{3}}$

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The set S={1,2,3−−−−−−−12} is to be portioned into three sets A, B, C of equal size. Thus A∪B∪C=S, A∩B=B∩C=A∩C=ϕ, the number of ways to partition S is

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