Q:

how many $4$ digit numbers are there with $3$ distinct digits, using all $3$ of them.

My question is, if I have for example digits $\{1,2,3\}$, how many $4$ digit numbers can I obtain if I have to use all of them in each combination?
Correct combinations:
$\{1,1,2,3\}$
$\{1,1,3,2\}$
$\{3,2,1,2\}$
$\{2,2,3,1\}$
Wrong combinations:
$\{1,1,2,1\}$ There is not digit $\{3\}$
$\{1,1,3,3\}$ There is not digit $\{2\}$
$\{2,2,2,3\}$ There is not digit $\{1\}$
And which would be the probability of this happening?
I guess that finding the combinations all I have to do is to divide this result by the total number of possible combinations.

A:

Now for $2$ digits in $4$ places we first put the $2$ digits in any $2$ places in only $1$ way( since we have to place both of them).Now there are $2$ places which are still empty.These two places can be filled in two different ways 
$(1)$ We can place any $1$ digit twice like $\{1222\}\{1211\}...$
$(2)$ We can place $2$ digits at the $2$ places like $\{1212\}\{1221\}$ and so on.
For $(1)$ the number of such numbers is $2 \cdot (4!/3!)=8$ (you have calculated) 
For $(2)$ the number of such numbers is $4!/(2! \cdot 2!) =6$
So adding $(1)$ and $(2)$ we get the total number of such numbers i.e $8+6=14$
So the probability is $14/2^4=0.875$.
So you should be very careful how we can fill those empty places after placing the digits. While calculating you need to consider all possible ways of filling those empty places(here you missed out the possibility $(2)$). Hope you have got your answer.

