Question

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

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

Answer 1

The correct option is C

$\frac{12!}{{\left(4!\right)}^3}$

Explanation for the correct option

Given set $S=\{1,2,3,..........,12\}$

Number of elements in the set$=12$

Also, $A\cup B\cup C=S$

Since, $A\cap B=B\cap C=A\cap C=\varnothing$ , all the elements in the set $A,B,C$ are distinct and also $A,B,C$ has equal size, so

1. Number of elements in $A=4$
2. Number of elements in $B=4$
3. Number of elements in $C=4$

Thus, number of ways to partition $S$ is given by

$$\begin{gathered} {}^{12}C_4\times {}^{8}C_4\times {}^{4}C_4=\frac{12!}{4!\times 8!}\times \frac{8!}{4!\times 4!}\times \frac{4!}{4!\times 0!}\left[\because {}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =\frac{12!}{{\left(4!\right)}^3} \end{gathered}$$

Hence, the correct option is option(C).