# no. of ways of $\displaystyle S = \{1,2,3,4,5,6,…,12\}$ is partitioned into three sets $A,B,C$ of equal size

Let $\displaystyle S = \{1,2,3,4,5,6,.....,12\}$ is partitioned into three sets $A,B,C$ of equal size

such that $A\cup B\cup C = S$ and $A\cap B\cap C = \phi.$ Then no. of ways to partition $S$ is

$\bf{My\; Solution::}$ Here we have to divide $12$ distinct object into $3$ group of equal size.

So no. of ways of forming a Group is $\displaystyle = \frac{\binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3}}{4!} = \frac{12!}{(4!)^3\cdot 3!}$

But Answer given as $\displaystyle = \frac{12!}{(4!)^3}$

Here i did not understand why we multiply answer by $4!$, because we have to partitioned into $3$

group and each contain $4$ elements(Where order has no importance.)

So plz explain me that step in Detail.

Thanks

Thanks Robjon and orangeskid , I have edited it.

• I suppose that exchanging e.g. $A,B$ counts extra. Otherwise see there. – ccorn Oct 6 '14 at 20:47

This is the same as asking how many ways you can arrange $4A$s, $4B$s, and $4C$s. $$\frac{12!}{4!\,4!\,4!}$$ For example: $$\begin{array}{c} 1&2&3&4&5&6&7&8&9&10&11&12\\ A&B&B&A&C&B&C&A&B&C&C&A \end{array}$$ represents $A=\{1,4,8,12\}$, $B=\{2,3,6,9\}$, and $C=\{5,7,10,11\}$.
I have no idea why the answer is given as $\dfrac{12!}{(3!)^4}$ rather than $\dfrac{12!}{(4!)^3}$.
$$\frac{12!}{(4!)^3 \cdot 3!} = 5775$$