Let $A_i$ be the event that we obtain exactly $k$ $i$'s (assuming that the dice are labeled $1,2,3,\ldots,m$ on their sides). Now inclusion-exclusion gives
$P(A_1 \cup A_2 \cup \ldots \cup A_m) = \sum P(A_i) - \sum P( A_i \cap A_j) + \sum P(A_i \cap A_j \cap A_k ) - \cdots $
where the summations range over distinct $i,j,k,\ldots .$
We have
$P(A_i) = { n \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-k} $
and
$P(A_i \cap A_j) = { n \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-k} \cdot { n-k \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-2k} $
and
$P(A_i \cap A_j \cap A_k)$ $ = { n \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-k} \cdot { n-k \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-2k} \cdot { n-2k \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-3k} $
and so on.
Simplifying these expressions, the probability that we get exactly $k$ of a number is given by
$P(A_1 \cup A_2 \cup \ldots \cup A_m) $
$ = { m \choose 1 } { n \choose k} \left( \frac{1}{m} \right)^k \left( 1 - \frac{1}{m} \right)^{n-k} $
$ - { m \choose 2 } { n \choose k} { n-k \choose k} \left( \frac{1}{m} \right)^{2k} \left( 1 - \frac{1}{m} \right)^{2n-3k}$
$ + { m \choose 3} { n \choose k} { n-k \choose k} { n-2k \choose k} \left( \frac{1}{m} \right)^{3k} \left( 1 - \frac{1}{m} \right)^{3n-6k} - \cdots $