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!?
 A: The calculation in the numerator counts all arrangements in which the pair comes first. That means that it counts the following six hands separately, even though they’re the same poker hand:
$$\begin{align*}
&\spadesuit 3,\clubsuit 3,\heartsuit 5,\clubsuit 8,\diamondsuit 9\\
&\spadesuit 3,\clubsuit 3,\heartsuit 5,\diamondsuit 9,\clubsuit 8\\
&\spadesuit 3,\clubsuit 3,\diamondsuit 9,\heartsuit 5,\clubsuit 8\\
&\spadesuit 3,\clubsuit 3,\diamondsuit 9,\clubsuit 8,\heartsuit 5\\
&\spadesuit 3,\clubsuit 3,\clubsuit 8,\heartsuit 5,\diamondsuit 9\\
&\spadesuit 3,\clubsuit 3,\clubsuit 8,\diamondsuit 9,\heartsuit 5\\
\end{align*}$$
Each of the $3!$ permutations of the singletons gets counted separately: the factors of $12,11$, and $10$ just guarantee that each singleton differs in rank from the pair and from the earlier singletons. The pair, however, is uniquely accounted for by the $13\binom42$, which chooses its rank and then two suits; there’s no other pair that could change places with it.
If you were counting hands with two pairs, you’d have
$$\frac{13\binom42\cdot12\binom42\cdot11\binom41}{2!}\;,$$
where the $2!$ accounts for the fact that the same two pairs could be listed in either order, while the singleton can’t change places with anything else. Here the calculation in effect lists the pairs first and then the singleton but counts these two orders separately, even though they’re the same hand:
$$\begin{align*}
&\spadesuit 3,\clubsuit 3,\heartsuit 5,\diamondsuit 5,\clubsuit 8\\
&\heartsuit 5,\diamondsuit 5,\spadesuit 3,\clubsuit 3,\clubsuit 8\\
\end{align*}$$
A: 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. 
A: 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
