What is the probability of getting exactly k different values out of n dices Say I got $n$ m-faced dices, each dice has value 1 to $m$ on one of the sides.
Now after one throw, what is the probability of getting exactly $k$ different values ?
For eg, Let $n = 7, m = 5, k = 3$,
Then ${1,2,3,3,3,3,3}$ or $3,3,5,2,5,3,5$ is a desired result as both set consists 3 different values
 A: First choose the $k$ different values for a factor of
$${m\choose k}$$
and  arrange them  in order.  Next partition  the $n$  rolls into  $k$
non-empty subsets, for a factor of
$${n\brace k}.$$
Match the ordered dice values to the sets in
$$k!$$
ways. This makes for a probability of
$$\bbox[5px,border:2px solid #00A000]{
m^{-n} \times {m\choose k} \times {n\brace k} \times k!.}$$
Sum this to see that it is a probability distribution:
$$m^{-n} \sum_{k=1}^m {m\choose k} \times {n\brace k} \times k!
\\ = m^{-n} \sum_{k=1}^m {m\choose k} n! [z^n] (\exp(z)-1)^k.$$
With  $n\ge 1$  we may  include  the term  for  $k=0$ as  it does  not
contribute to $[z^n]$ to get
$$m^{-n} n! [z^n] \sum_{k=0}^m {m\choose k}  (\exp(z)-1)^k
\\ = m^{-n} n! [z^n] \exp(zm)
= 1$$
which confirms it being a probability distibution. We can also compute
the expectation which requires
$$m^{-n} \sum_{k=1}^m {m\choose k} n! [z^n] k (\exp(z)-1)^k
\\ = m^{-n} n! [z^n] \sum_{k=1}^m {m\choose k}  k (\exp(z)-1)^k
\\ = m^{-n} n! [z^n] m \sum_{k=1}^m {m-1\choose k-1}  (\exp(z)-1)^k
\\ = m^{-n} n! [z^n] m (\exp(z)-1)
\sum_{k=1}^m {m-1\choose k-1}  (\exp(z)-1)^{k-1}
\\ = m^{-n} n! [z^n] m (\exp(z)-1) \exp((m-1)z)
\\ = m^{-n} n! [z^n] m (\exp(mz)-\exp((m-1)z))
\\ = m \times \left(1 - m^{-n} (m-1)^n\right).$$
We get
$$\bbox[5px,border:2px solid #00A000]{
m \times \left(1 - \left(1-\frac{1}{m}\right)^n\right).}$$
There  is  also this  very  basic  Maple  code that  implements  these
statistics.

with(combinat);

ENUM :=
proc(n, m)
option remember;
local d, mset, idx, gf;

    gf := 0;

    for idx from m^n to 2*m^n-1 do
        d := convert(idx, base, m);
        mset := convert(d[1..n], `multiset`);
        gf := gf + u^nops(mset);
    od;

    gf;
end;

GFX := (n, m) ->
add(u^k*binomial(m,k)*stirling2(n,k)*k!, k=1..m);

PROB := (n,m) -> subs(u=1, GFX(n, m)) / m^n;
EXPT := (n,m) -> subs(u=1, diff(GFX(n, m), u)) / m^n;

EXPT2 := (n, m) -> m*(1 - (1-1/m)^n);

Remark. User  @orlp suggests looking at  the generating function.
Recall    that     for    Stirling    numbers    of     the    second
kind
we have
$$\sum_{n\ge 0} {n\brace k} x^n
= \prod_{r=1}^k \frac{x}{1-rx}.$$
We get for the generating function
$$G_k(z) = \sum_{n\ge 1} z^n \times
m^{-n} \times {m\choose k} \times {n\brace k} \times k!
\\ = \frac{m!}{(m-k)!} \sum_{n\ge 1} m^{-n} z^n
[x^n] \prod_{r=1}^k \frac{x}{1-rx}
\\ = \frac{m!}{(m-k)!} \prod_{r=1}^k \frac{z/m}{1-rz/m}
\\ = \frac{m!}{(m-k)!} \prod_{r=1}^k \frac{z}{m-rz}.$$
This may be re-written as
$$m^{\underline{k}} \prod_{r=1}^k \frac{1}{m/z-r}
= m^{\underline{k}} \prod_{r=0}^{k-1} \frac{1}{m/z-1-r}
= \frac{m^{\underline{k}}}{(m/z-1)^{\underline{k}}}.$$

G1 := (m, k, mx) ->
add(z^n*binomial(m,k)*stirling2(n,k)*k!/m^n, n=0..mx);

FF := (val, q) -> mul(val-p, p=0..q-1);

G2 := (m, k, mx) ->
convert(series(FF(m,k)/FF(m/z-1,k), z=0, mx+1), polynom);

A: The equi-probable space to consider is that resulting from considering the dice labelled, i.e. by throwing a die $n$ times and considering the 
results in sequence.
Call $N(k,m,n)$ the number of ways to obtain exactly $k$ distinct outcomes, from a $m$-face die , in $n$ throws.
Let's start from throwing the first die $n=1$: we can have just $k=1$ (distinct) outcome, and we can obtain that in $m$ ways.   
Throwing the second time, either we get the same result as before ($1$ way) or we get a different result ($m-1$ ways).
The first case will contribute to $N(1,m,2)$, the second to $N(2,m,2)$.
If upon the $n-1$-th throw we totaled $k$ distinct outcomes, then with the throw $n$ we have $k$ ways to contribute to N(k,m,n) 
and $m-k$ ways to contribute to $N(k+1,m,n)$.
This leads to the following recurrence
$$ \bbox[lightyellow] {  
N\left( {k,m,n} \right) = k\,N\left( {k,m,n - 1} \right) + (m - k + 1)N\left( {k - 1,m,n - 1} \right) + \left[ {0 = k = n} \right]
 }$$
where $[P]$ denotes the Iverson bracket
The solution to the recurrence is
$$ \bbox[lightyellow] {  
N\left( {k,m,n} \right) = \left\{ \matrix{
  n \cr 
  k \cr}  \right\}m^{\,\underline {\,k\,} }  = k!\left\{ \matrix{
  n \cr 
  k \cr}  \right\}\left( \matrix{
  m \cr 
  k \cr}  \right)
 }$$
because in fact
$$
\eqalign{
  & N\left( {k,m,n} \right) = \left\{ \matrix{
  n \cr 
  k \cr}  \right\}m^{\,\underline {\,k\,} }  = k\,\left\{ \matrix{
  n - 1 \cr 
  k \cr}  \right\}m^{\,\underline {\,k\,} }  + (m - k + 1)\left\{ \matrix{
  n - 1 \cr 
  k - 1 \cr}  \right\}m^{\,\underline {\,k - 1\,} }  =   \cr 
  &  = \left( {k\,\left\{ \matrix{
  n - 1 \cr 
  k \cr}  \right\} + \left\{ \matrix{
  n - 1 \cr 
  k - 1 \cr}  \right\}} \right)m^{\,\underline {\,k\,} }  = \left\{ \matrix{
  n \cr 
  k \cr}  \right\}m^{\,\underline {\,k\,} }  \cr} 
$$
And we reach, by another way, to the same conlusion as that already given by Marko Riedel.
Note that ${m \choose k}$ is the number of ways to select $k$ elements
out of $n$, and $k!\left\{ \matrix{  n  \cr k \cr}  \right\}$ is the number of surjections
from a set with $n$ elements to a set with $k$ elements.
A: This answer isn't as useful as Marko Riedel's, but I found the generating function. We have
$$f(n, k, m) = \frac{m-(k-1)}{m}f(n-1, k-1, m) + \frac{k}{m}f(n-1, k, m)$$
If we fix $m$ we get $p(n,k)$. Let's write a power series for $p(n,k)$:
$$A_k(x) = \sum_{n=1}^\infty p(n,k)x^n$$
But we know that
$$A_k(x) = p(1, k)x + \sum_{n=2}^\infty \left( \frac{m-(k-1)}{m}p(n-1, k-1) + \frac{k}{m}p(n-1, k)\right)x^n$$
Let's assume $k > 1$ for the above formula, giving $p(1, k) = 0$.
$$A_k(x) = \frac{m-(k-1)}{m}\sum_{n=2}^\infty  p(n-1, k-1)x^n +\frac{k}{m}\sum_{n=2}^\infty p(n-1, k)x^n$$
$$A_k(x) = \frac{m-(k-1)}{m}\sum_{n=1}^\infty  p(n, k-1)x^{n+1} +\frac{k}{m}\sum_{n=1}^\infty p(n, k)x^{n+1}$$
$$A_k(x) = \frac{m-(k-1)}{m}\cdot xA_{k-1}(x) +\frac{k}{m}\cdot xA_k(x)$$
$$(m- kx)A_k(x) = (m-(k-1))\cdot xA_{k-1}(x) $$
$$A_k(x) = x\frac{m-k+1}{m - kx}A_{k-1}(x) $$
And it should be easy to see that $p(n, 1) = m^{1-n}$, giving:
$$A_1(x) = \sum_{n=1}^\infty m^{1-n}x^n = \frac{mx}{m-x}$$
Giving the final generating function:
$$A_k(x) = \frac{mx}{m - x}\prod_{i=2}^k x\frac{m-k+1}{m-kx} = \frac{(-m)_k}{(1-\frac{m}{x})_k}$$
where $(x)_n$ is the Pochhammer symbol.
A: For $i=1,\dots,m$ let $X_i$ take value $1$ if side $i$ shows up and let it take value $0$ otherwise. Then you are looking for $$P(X_1+\cdots X_m=k)$$
Based on symmetry we find that this equals $$\binom{n}{k}P\left(X_1=\cdots=X_k=1\wedge X_{k+1}=\cdots=X_m=0\right)$$
and also: $$\binom{n}{k}P\left(X_1=\cdots=X_k=1\mid X_{k+1}=\cdots=X_n=0\right)P(X_{k+1}=\cdots=X_m=0)$$
It is not difficult to find that $P(X_{k+1}=\cdots=X_m=0)=\left(\frac{m-k}{m}\right)^{n-k}$.
It remains to find the conditional probability which is actually the answer on the following question:

"If  $n$ dice having $k$ faces are thrown then what is the probability that all faces show up?"

This is a problem that can be solved with the inclusion/exclusion principle and symmetry.
From here we deal with this new question and for $=1,\dots,k$ let $A_i$ denote the event that face $i$ does not turn up.
Then to be found is: $$1-P(A_1\cup\cdots\cup A_k)$$
First give it a try yourself.
A: When I jumped on this question I figured it was not too difficult and some hints would be helpful (see the second section). But now that section is more humorous than useful, as it is really just a 'stream-of-consciousness' display.
With the wonderful answers found here, the counting argument can also be easily 'reversed engineered' (see G Cab's concluding remarks). 
The expressions $\left\{ \matrix{  n  \cr k \cr}  \right\}$ are the Stirling numbers of the second kind and count the number of different equivalence relations with precisely ${\displaystyle k}$ equivalence classes that can be defined on an ${\displaystyle n}$ element set. 
Proposition 1: Let $X$ be a finite set with $n$ elements and $Y$ be a finite set with $k$ elements. Then there are $k!\left\{ \matrix{  n  \cr k \cr}  \right\}$ surjective mappings of $X$ onto $Y$.
Proof
To start, observe that if $f$ is any such surjection, and $\sigma$ is a bijective transformation (permutation) of $Y$ not equal to the identity map, then $\sigma \circ f$ is a different surjective map than $f$. But $f$ also defines an equivalence relation, $\{f^{-1}(y_0)\}$, on $X$ so that $f$ 'factors thru' the corresponding quotient set. It is simple task showing that for any surjection $f$, $\sigma_1 \circ f = \sigma_2 \circ f$ if and only if $\sigma_1 = \sigma_2$. If is also easy to check that any other surjection with the same quotient set as $f$ has the form $\sigma^{`} \circ f$ for some (unique) permutation $\sigma^{`}$. Using the rule of product gives us the desired result.
$\blacksquare$
Proposition 2: Let $X$ be a finite set with $n$ elements and $Y$ be a finite set with $m$ elements. Then there are $ {m \choose k} k!\left\{ \matrix{  n  \cr k \cr}  \right\}$ mapping of $X$ into $Y$ that 'hit' exactly $k$ elements.
Proof
This follows immediately from proposition 1 and, again, by applying the rule of product. $\qquad \blacksquare$

Counting Hints:
Notation: $\bar j = \{1,2,\dots,j \}$.
There are $m^n$ functions mapping $\bar n$ into $\bar m$.
How many surjective maps are there from $\bar n$ onto $\bar k$?
$\quad$ (this is a 'warm up' exercise).
How many partitions are there of $n$ into $k$ parts?
${\displaystyle m^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}}$
For your example, $n = 7, m = 5, k = 3$, we want to count a class of functions including $(1,2,3,3,3,3,3)$. So, $7 = 5 + 1 + 1$, and
$5 + 1 + 1 + 0 + 0 $
$5 + 1 + 0 + 1 + 0 $
$5 + 1 + 0 + 0 + 1 $
$5 + 0 + 1 + 1 + 0 $
$5 + 0 + 1 + 0 + 1 $
$5 + 0 + 0 + 1 + 1 $
etc.
So, $(5) (6) = (5) {4 \choose 2} = 30$ and
$\tag 1 30 {7 \choose 5,1,1,0,0}$
would 'pick up' $(1,2,3,3,3,3,3)$, since it contributes a $1\text{-count}$ to ${7 \choose 1,1,5,0,0}$.
A: The probability of $n$ $m$-sided dice showing exactly $k$ values.
Let $S(i)$ be the set of rolls of $n$ $m$-sided dice where face $i$ does not show up. Then
$$
\begin{align}
N(j)
&=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|\\
&=\binom{m}{j}(m-j)^n\tag1
\end{align}
$$
The Generalized Inclusion-Exclusion Principle says that the number of rolls in exactly $m-k$ of the $S(i)$ is
$$\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
\begin{align}
\sum_{j=0}^m(-1)^{j-m+k}\binom{j}{m-k}N(j)
&=\sum_{j=0}^m(-1)^{j-m+k}\binom{j}{m-k}\binom{m}{j}(m-j)^n\tag{2a}\\
&=\sum_{j=0}^m(-1)^{j-m+k}\binom{m}{m-k}\binom{k}{j-m+k}(m-j)^n\tag{2b}\\
&=\sum_{j=0}^k(-1)^{k-j}\binom{m}{k}\binom{k}{j}\,j^n\tag{2c}\\
&=\sum_{j=0}^k\sum_{i=0}^n(-1)^{k-j}\binom{m}{k}\binom{m}{j}\binom{j}{i}\stirtwo{n}{i}\,i!\tag{2d}\\
&=\sum_{j=0}^k\sum_{i=0}^n(-1)^{k-j}\binom{m}{k}\binom{k}{i}\binom{k-i}{j-i}\stirtwo{n}{i}\,i!\tag{2e}\\
&=\sum_{j=0}^{k-i}\sum_{i=0}^n(-1)^{k-j-i}\binom{m}{k}\binom{k}{i}\binom{k-i}{j}\stirtwo{n}{i}\,i!\tag{2f}\\
&=\sum_{i=0}^n(-1)^{k-i}\binom{m}{k}\binom{k}{i}[i=k]\stirtwo{n}{i}\,i!\tag{2g}\\
&=\binom{m}{k}\stirtwo{n}{k}\,k!\tag{2h}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $(1)$
$\text{(2b)}$: $\binom{j}{m-k}\binom{m}{j}=\binom{m}{m-k}\binom{k}{j-m+k}$
$\text{(2c)}$: substitute $j\mapsto m-j$
$\text{(2d)}$: $j^n=\sum\limits_{i=0}^n\binom{j}{i}\stirtwo{n}{i}\,i!$
$\text{(2e)}$: $\binom{m}{j}\binom{j}{i}=\binom{k}{i}\binom{k-i}{j-i}$
$\text{(2f)}$: substitute $j\mapsto j+i$
$\text{(2g)}$: $\sum\limits_{j=0}^{k-i}(-1)^j\binom{k-i}{j}=[i=k]$ using Iverson brackets
$\text{(2h)}$: apply the Iverson brackets

Thus, the probability of rolling exactly $n$ $m$-sided dice and showing exactly $k$ values is
$$
\bbox[5px,border:2px solid #C0A000]{\frac1{m^n}\binom{m}{k}\stirtwo{n}{k}\,k!}\tag3
$$
