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!?

The probability shouldnt be 0.42256
To have the total sum of combinations of five cards, all five cards would have to be different, which is selecting without replacement. We could either compute 52 × 51× 50 × 49 × 48 or notice that this is the same as the permutation 52P5 = 311,875,200.
Probability= Combinations of hands with one pair/ Total number of permutations
Probability=0.003521
