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I am struggling with the following question:

How many ways are there to divide $k$ distinct balls into $n$ distinct cells where the order of the balls in the cell is relevant and each cell has the same nunber of balls.

Algebraically, I am getting $k!$ ,But this seems wrong to me and I am not sure I am using the correct formulas.

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  • $\begingroup$ You might wanna show what was your line of thought when you arrived at $k!$, this would make the folks here address your concern more precisely. $\endgroup$ – user350331 Dec 31 '16 at 12:23
  • $\begingroup$ What do you mean "each cell has the same number of balls"? Do you mean that $k$ is a multiple of $n$, say $k=n\times m$ and each cell gets exactly $m$? If so, then pick $m$ for the first cell, so $\binom {mn}m$, then $m$ for the second, $\binom {m(n-1)} m$ and so on and multiply. $\endgroup$ – lulu Dec 31 '16 at 12:24
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This means that $k$ divides $n$. If the order of balls was unimportant, you'd have $\binom{k}{\frac{n}{k}} \binom{k-\frac{k}{n}}{\frac{k}{n}} \ldots \binom{\frac{k}{n}}{\frac{k}{n}}$. Since the order of balls in the bin is important, you need to multiply by $(\frac{k}{n})^n$ to get $k!$

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Obviously, $\frac{k}{n}$ has to be an integer $=z$,say.

The simplest way to conceive it is to take each of the $k!$ permutations of the balls placed in a row, and just put dividers after every $z$ balls.

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