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The set $S = \left\{ {1,2,3, \ldots ,12} \right\}$ is to be partitioned into three sets $A,\,B,\, C$ of equal size . Thus $A \cup B \cup C = S$ અને $A \cap B = B \cap C = C \cap A = \emptyset $ . The number of ways to partition $S$ is
$\frac{{12!}}{{{{\left( {4!} \right)}^3}}}$
$\;\frac{{12!}}{{{{\left( {4!} \right)}^4}}}$
$\;\frac{{12!}}{{3!{{\left( {4!} \right)}^3}}}$
$\;\frac{{12!}}{{3!{{\left( {4!} \right)}^4}}}$
Solution
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 $=^{12} C_{4} \times^{8} C_{4} \times^{4} C_{4}$
$=\frac{12 !}{8 ! 4 !} \times \frac{8 !}{4 ! 4 !} \times \frac{4 !}{0 ! 4 !}=\frac{12 !}{(4 !)^{3}}$
