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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.

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  • $\begingroup$ 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
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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!

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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.

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