Combinatorics question. Probability of poker hand with one pair If we assume that all poker hands are equally likely, what is the probability of getting 1 pair? 
So the solution is
 
I understand numerator part, but I do not understand why in denominator we have 3!.
In the books it says "The selection of the cards “b”, “c”, and “d” can be permuted in any of the 3! ways and the same hand results."
But we also have one pair, e.g (a,a,b,c,d) and as I understand it makes no difference how our cards are arranged. So why we are not dividing by 5!?

We count the one pair hands in a somewhat different way. The kind we have a pair in can be chosen in $\binom{13}{1}$ ways. For every choice of kind, the actual cards can be chosen in $\binom{4}{2}$ ways.
Now we choose the $3$ kinds that we have singletons in. These can be chosen in $\binom{12}{3}$ ways. Imagine arranging these kinds in increasing order. The card that represents the smallest kind chosen can be chosen in $\binom{4}{1}$ ways. Now the second highest kind can have its card chosen in $\binom{4}{1}$ ways, and so can the third kind. This gives a total of 
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3.$$
Note that $\binom{12}{3}=\frac{(12)(11)(10)}{3!}$. This matches the expression you were asking about. 
