Alright so if i have

n! Is the number of permutations assuming all objects are distinguishable

and we want m of the n objects indistinguishable so I think I would do something of the form

$n_1$ indistinguishable objects of type 1

$n_2$ indistinguishable objects of type 2


$n_k$ indistinguishable objects of type k

but since the total of indistinguishable is supposed to be m

m = $n_1$ + $n_2$ + ... + $n_k$

I think, but I'm not sure what $n_1$ etc.. would be

maybe just



It is advantageous to consider a more general problem. Assume there are $k$ types of objects, each kind $i$ having $n_i$ indistinguishable representatives, the overall number of objects being $n=\sum_{i=1}^k n_i$.

Then the overall number of possible permutations is determined by the multinomial coefficient: $$ \frac{n!}{\prod_{i=1}^k n_i!}. $$

Observe that this expression remains valid also when some $n_i$ are $0$.

  • $\begingroup$ So just to make sure I understand, it would be n! / $n_1$! $n_2$! ... $n_m$! ? Sorry if I misunderstand, I'm just not familiar with what your proposing. $\endgroup$ – Brownie Apr 29 '19 at 21:31
  • 1
    $\begingroup$ @Brownie Yes, it is (in terms of your question the last index should be $k$). $\endgroup$ – user Apr 29 '19 at 21:33

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