Probability of 2 sets of triples in a 6 card hand The deck is a standard 52 card deck. My solution was:
The first card drawn can be anything, so 52 possible cards. The next card has to be the same value, so there are 3 possible cards. The last card can only be from the 2 remaining cards. The next set of triples can start from one of 48 cards since the last of the first value is dead to the problem, then once again there are 3 then 2 possibilities. I believe that covers all possible hands with double triples. The number of possible hands is 52 choose 6, so 52!/(6!46!). Simplified, that should make the probability equal:
(52x48x2x2x3x3)(6!46!)/52!
I can't help but feel like I messed something up. Do I need to square the 6! to account for the orders of the hand, something else, or is my answer correct?
 A: A "favourable" hand can be selected as follows:


*

*choose two different values from $13$, order not important. . . . . $C(13,2)$ ways;

*choose three of the four cards from each of these values. . . . . $C(4,3)^2=4^2$ ways.


Since the total number of hands is $C(52,6)$, the required probability is
$$\frac{C(13,2)4^2}{C(52,6)}\ .$$
The problem with your solution is that you have chosen the cards forming triples in a particular order: for example you have chosen
$\def\\#1{{\rm#1}}$
$$\heartsuit2,\,\clubsuit2,\,\spadesuit2,\,
  \diamondsuit \\Q,\,\heartsuit \\Q,\,\clubsuit \\Q\quad\hbox{and}\quad
  \clubsuit \\Q,\,\diamondsuit \\Q,\,\heartsuit \\Q,\,
  \heartsuit2,\,\spadesuit2,\,\clubsuit2$$
as two different hands, when they are actually the same hand.
A: The two kinds that we can have $3$ of can be chosen in $\binom{13}{2}$ ways. Given the kinds, the actual $3$ cards of each kind can be chosen in $\binom{4}{3}\binom{4}{3}$ ways, for a total of $\binom{13}{2}\binom{4}{3}^2$.  
For the probability, divide by $\binom{52}{6}$. 
Remark: In the attempted solution, there is multiple counting going on. You are counting getting $2$ of spades, $2$ of hearts, $2$ of diamonds as different from getting $2$ of diamonds, $2$ of spades, $2$ of hearts. 
You are also counting three Kings and  three Jacks as different from three Jacks and three Kings. 
If in the denominator you have the number of hands, then the numerator must also count hands.   
But the multiple counting is fixable. If you recognize that you have counted the possibilities for three $2$'s using order, you can divide your answer by $3!$ to get rid of that multiple counting. We divide by another $3!$ for the other kind. And we divide by $2!$ to get rid of the double-counting of kinds. So your answer, after division by $3!3!2!$, becomes correct. 
Indeed, one can use deliberate multiple counting as a strategy, which can work nicely if one can get a handle on the exact amount of multiple counting. 
