Question

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

• A. $\frac{12!}{3!(4!{)}^3}$
• B. $\frac{12!}{3!(3!{)}^4}$
• C. $\frac{12!}{(4!{)}^3}$
• D. $\frac{12!}{(3!{)}^4}$

Answer 1

The correct option is C $\frac{12!}{(4!{)}^3}$

There are $12$ elements.

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

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${=}^{12}{C}_4{\times }^8{C}_4{\times }^4{C}_4$

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