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 Robjon and orangeskid , I have edited it.

  • $\begingroup$ I suppose that exchanging e.g. $A,B$ counts extra. Otherwise see there. $\endgroup$ – 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 $\dfrac{12!}{(4!)^3}$.


$$\frac{12!}{(4!)^3 \cdot 3!} = 5775$$

  • 1
    $\begingroup$ A few words of explanation would surely improve this Answer, esp. as the OP has given enough context that the point at which their computation went off the rails is fairly clear. $\endgroup$ – hardmath Oct 6 '14 at 21:36

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