$\begin{array}{1 1}\large\frac{12!}{(3!)^5} \\ \large\frac{12!}{(4!)^3} \\\large\frac{12!}{(3!)^4} \\ \large\frac{12!}{3!(4!)^3} \end{array}$

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- $^nC_r=\large\frac{n!}{(n-r)!r!}$
- $0!=1$

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

There are 12 elements.

Three disjoint sets A,B and C of same size are to be formed.

$\Rightarrow $ Each set has 4 elements each.

No.of ways of selecting any 4 elements from 12 elements for set $A =^{12}C_4$.

No.of ways of selecting any 4 elements from 8 elements for set $B =^{8}C_4$.

No.of ways of selecting any 4 elements from 4 elements for set $C =^{4}C_4$.

Total no. of partitions$=^{12}C_4\times^{8}C_4\times^{4}C_4$

$=\large\frac{12!}{8!4!}\times\frac{8!}{4!4!}\times\frac{4!}{0!4!}=\frac{12!}{(4!)^3}$

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