Number of ways colour $n$ distinct objects with at most $n$ different colours without permutations of colours Suppose that I have $n$ distinct objects. In how many ways can I colour the $n$ objects with at most $n$ distinct colours up to permutations of colours. For e.g 
$1,1,1,1$ is the same thing as $2,2,2,2$
$1,1,2,3$ is the same thing as $4,4,1,2$
but $1,1,2,3$ is different from $2,3,1,1$
My approach is count the number of ways to use exactly $i \leq n$ colours to colour the $n$ objects using the inclusion-exclusion principle then divide by $ \binom{n}{i}$. Then I add everything up. However I'm not sure if the method is correct and I'm also looking for a general solution in terms on $n$.
 A: What you want to count is the number of different partitions of an $n$ element set into $\leq n$ nonempty subsets. The number of these partitions is the Bell number $B_n$. There is a recursive formula for these and an asymptotic formula involving the Lambert $W$-function.
A: By way of enrichment  I would like to point out that  this can also be
done  by  Power  Group  Enumeration   as  defined  by  Harary  and
Palmer. Here we  have that the slot permutation group  is the identity
$E_n$ with  cycle index  $Z(E_n) = a_1^n$  and the  object permutation
group  is the  symmetric group  $S_n$ with  cycle index  given by  the
exponential formula
$$Z(S_n) = [z^n] \exp\left(\sum_{l\ge 1} a_l \frac{z^l}{l}\right).$$
Now to apply PGE we must cover the single permutation $a_1^n$ with
cycles from  the permutations in  $Z(S_n).$ Clearly if the  latter has
$k$ fixed  points this can be  done in $k^n$ ways.  Hence we introduce
the mixed generating function with fixed points marked
$$G(z, u) = \exp\left(uz-z+\log\frac{1}{1-z}\right).$$
The answer is thus given by (we turn $u^k z^n$ into $k^n z^n$)
$$ [z^n] \left. \left(u\frac{\partial}{\partial u}\right)^n G(z, u)
\right|_{u=1}
\\ =  [z^n] \left. \left(u\frac{\partial}{\partial u}\right)^n
\frac{1}{1-z} \exp(z(u-1))
\right|_{u=1}
\\ =   \left. \left(u\frac{\partial}{\partial u}\right)^n
\sum_{q=0}^n \frac{(u-1)^q}{q!}
\right|_{u=1}
\\ =   \left. \left(u\frac{\partial}{\partial u}\right)^n
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} u^p
\right|_{u=1}
\\ = \sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} p^n.$$
We recognize Stirling  numbers at this point but we  may also continue
with the EGF
$$F(w) = \sum_{n\ge 0} \frac{w^n}{n!}
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} p^n
\\ = \sum_{n\ge 0} \frac{w^n}{n!}
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} n! [v^n] \exp(pv)
\\ = \sum_{n\ge 0} w^n [v^n]
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} \exp(pv)
\\ = \sum_{n\ge 0} w^n [v^n]
\sum_{q=0}^n \frac{1}{q!}
(\exp(v)-1)^q.$$
Now since $\exp(v)-1=v+\cdots$  we may extend $q$ beyond  $n$ as there
is no contribution to the  coefficient extractor $[v^n]$ in that case,
getting
$$\sum_{n\ge 0} w^n [v^n]
\sum_{q\ge 0} \frac{1}{q!}
(\exp(v)-1)^q
\\ = \sum_{n\ge 0} w^n [v^n] \exp(\exp(v)-1)
= \exp(\exp(w)-1)$$
and we see  that we indeed have  Bell numbers here as  observed in the
reply that  was first to appear.  Note also that PGE  rests on the
Burnside lemma which was suggested in the comments.
