How many ways in which $m\cdot n$ distinct objects can be divided equally into $n$ groups?

The answer is $$\frac{(mn)!}{(m!)^n n!}$$

Can someone please supply the intuition behind this answer?

Thanks in advance.

  • $\begingroup$ Start by thinking about the numerator. Once you understand it, think about the denominator. In that order. In the denominator, think about the $m!$ before you think about the $\vphantom(^n$. $\endgroup$ – dfeuer Dec 27 '13 at 5:01
  • $\begingroup$ Just for completeness: This question is related to yours and has a number of answers that may be interesting for you as well. $\endgroup$ – jpvee Nov 27 '17 at 8:37

Imagine groups are written down in a row. This is same as permuting the original $n\cdot m$ objects and assigning the first $m$ object to the first group, the second $m$ objects to second group etc.

Now each such group has $m!$ ways it can be permuted, so there are $(m!)^n$ permutations that give the same groups.

Hence the answer $$ \frac{(m\cdot n)!}{(m!)^n} $$

You need to fill in the gaps!


First, let us consider an easy example.

If you want to divide $9$ distinct balls in the $A, B, C$ boxes, the answer is $$\binom{9}{3}\times\binom{6}{3}.$$ First choose three balls for $A$, then choose three balls for $B$.

On the other hand, if you want to divide $9$ distinct balls into three name-less boxes, the answer is $$\frac{\binom{9}{3}\times\binom{6}{3}}{3!}.$$

Now, let us come back to your question.

Your group has no names, so the answer is $$\frac{\binom{mn}{m}\times\binom{mn-m}{m}\cdots\times\binom{2m}{m}}{n!}.$$

This is equal to what you wrote.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.