<s>[INST] Probability of $2$ pairs on $5$ dice We throw $5$ dice. What is the probability of getting $2$ pairs ?
My solution says it is $$\frac{6\cdot 5\cdot 4\cdot 5! }{6^5\cdot 2\cdot 2!\cdot 2!},$$ where as for me it's $$\frac{6\cdot 5\cdot 4\cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
I do as follows: Throwing $5$ dice is the same thing as throwing one die $5$ times. What we want is $AABBC$.
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $\frac{5!}{2!2!}$. At the end, I get
$$\frac{6\cdot 5\cdot 4 \cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
What's wrong in my argument?
 [/INST] Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-\dfrac16=\dfrac56$
Taking your approach, you would say the probability should be $\dfrac{6\cdot 5 \cdot 2!}{6^2\cdot 1!\cdot 1!} = \dfrac{10}{6}$, which is clearly wrong.  The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)  


*

*You could say there are ${6 \choose 2}=\dfrac{6 \times 5}{2}$ ways of choosing $A$ and $B$

*Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
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