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# If the set $S={1,2,3...........12}$ is to be partitioned into $3$ sets $A,B,C$ of equal size so that $A\cup B\cup C=S$ and $A\cap B=B\cap C=C\cap A=\phi$, then the no. of ways the partition can be done is

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

Can you answer this question?

Toolbox:
• $^nC_r=\large\frac{n!}{(n-r)!r!}$
• $0!=1$
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$=^{12}C_4\times^{8}C_4\times^{4}C_4$
$=\large\frac{12!}{8!4!}\times\frac{8!}{4!4!}\times\frac{4!}{0!4!}=\frac{12!}{(4!)^3}$

answered Nov 6, 2013