We throw $5$ dice: What is the probability to have $4$ different numbers? We throw $5$ dice: What is the probability to have $4$ different numbers? 
I know it is $$\frac{6\cdot 5\cdot 4\cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $\binom{6}{4}$ numbers. Then I have $$\frac{6\cdot 5\cdot 4\cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways. 


*

*If I want all dice different, this argument works: I take $\binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $\frac{5\cdot 5\cdot 4\cdot 3\cdot 2}{5!}5!=6\cdot 5\cdot 4\cdot 3\cdot 2$ possibilities, that is the correct answer. 


So why doesn't it work with the previous situation ?

Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.  
Suppose the dice are five different colors so that we can distinguish between them.  There are $\binom{5}{2}$ ways for two of the five dice to display the same number and $\binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display.  The number of possible ways those three dice can display the numbers is $5 \cdot 4 \cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list).  Hence, the number of favorable outcomes is 
$$\binom{5}{2} \cdot 6 \cdot 5 \cdot 4 \cdot 3$$
from which we obtain the probability
$$\frac{\binom{5}{2} \cdot 6 \cdot 5 \cdot 4 \cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
