Card Hand Combinations A probability question asks how many ways one can select a 5 card one pair hand (that is a pair of cards are the same number and the other 3 are different) from a standard 52 card deck. 
The answer for the question from the back of my book was calculated as $$\frac{13\pmatrix{4\cr2}\cdot12\pmatrix{4\cr1}\cdot11\pmatrix{4\cr1}\cdot10\pmatrix{4\cr1}}{3!}$$ Why is  the denominator not $4!$ since there are 4 groups that we are trying to arrange?
How is this any different than if you have 5 letters a, b, c, d, e and you are trying to find the amount of ways you can get groups of 3 using $$\frac{5\cdot4\cdot3}{3!}$$ since there are 3 elements to be arranged so you need to divide out the 3! permutations an ordered triplet can produce?
Thanks
EDIT: Second example was worded wrong, corrected now
 A: I would solve the problem in a marginally different way. The kind you have $2$ of can be chosen in $\binom{13}{1}$ ways.  For each of these ways, the actual cards can be chosen in $\binom{4}{2}$ ways. So far the same as yours. Then the kinds you have $1$ each of can be chosen in $\binom{12}{3}$ ways.  Once you have chosen these kinds, line them up in order of strength. In that order, choose the actual cards. This can be done in $\binom{4}{1}^3$ ways, for a total of
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3.$$
This is the same answer as yours, with the division by $3!$ "hidden" in $\binom{12}{3}$.  
If, as you did, you use the product $\binom{12}{1}\binom{11}{1}\binom{10}{1}$, then every one of the $3$ useless card choices is counted a total of $3!$ ways. So we divide by $3!$. We do not divide by $4!$ because it is only the singleton cards that have been multiply counted. What has happened is a kind of hybrid counting, some without order (what we have $2$ of) and the rest in order. 
We could count in a way that forces division by $5!$ (note, not $4!$) by following the order of the dealing.  Choose the kind we have $2$ of. This can be done in $\binom{13}{1}$ ways. Now choose when these cards come to you. Their locations in the dealing can be chosen in $\binom{5}{2}$ ways. Now fill the leftmost such location with a card of the right kind. There are $\binom{4}{1}$ ways to do this. Then fill the second such location. this can be done in $\binom{3}{1}$ ways. Now fill the empty locations, from left to right. This can be done in $\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}\binom{10}{1}\binom{4}{1}$ ways, for a total of 
$$\binom{13}{1}\binom{5}{2}\binom{4}{1}\binom{3}{1}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}\binom{10}{1}\binom{4}{1}.$$
We have counted every hand $5!$ times, so finally we need to divide by $5!$. 
