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
$$P(A_1 \cup A_2 \cup \ldots \cup A_m) $$
$$ = { m \choose 1 } { n \choose k} \left( \frac{1}{m} \right)^k
\left( 1 - \frac{1}{m} \right)^{n-k} $$
$$ - { m \choose 2 } { n \choose k} { n-k \choose k} \left( \frac{1}{m} \right)^{2k}
\left( 1 - \frac{1}{m} \right)^{2n-3k}$$
$$ + { m \choose 3} { n \choose k} { n-k \choose k} { n-2k \choose k} \left( \frac{1}{m} \right)^{3k} \left( 1 - \frac{1}{m} \right)^{3n-6k} - \cdots $$