A) \[\frac{12!}{{{(4!)}^{3}}}\]
B) \[\frac{12!}{{{(4!)}^{4}}}\]
C) \[\frac{12!}{3!{{(4!)}^{3}}}\]
D) \[\frac{12!}{3!{{(4!)}^{4}}}\]
Correct Answer: A
Solution :
[a] Set \[S=\{1,2,3,...12\}\] \[A\cup B\cup C=S,A\cap B=B\cap C=A\cap C=\phi \] \[\therefore \] The number of ways to partition \[{{=}^{12}}{{C}_{4}}{{\times }^{8}}{{C}_{4}}{{\times }^{4}}{{C}_{4}}\] \[=\frac{12!}{4!8!}\times \frac{8!}{4!4!}\times \frac{4!}{4!0!}=\frac{12!}{{{(4!)}^{3}}}\]
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