# how many permutations can be formed from n objects when m of them are indistinguishable?

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

n!/m!

## 1 Answer

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 $$0$$.

• 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. – Brownie Apr 29 '19 at 21:31
• @Brownie Yes, it is (in terms of your question the last index should be $k$). – user Apr 29 '19 at 21:33