Cfrgtkky

I'm new to combinatorics, and I'm having difficulties with the following problem.

From standard deck of 52 cards, we want the number of "3 of a kinds"
that are not "4 of a kinds" and not "full houses."

Here's my solution, but it looks like I'm overcounting.

Step 1 - Choose kind: $13 choose 1$ = $13$ ways

Step 2 - Choose three of that kind: since there are 4 of each kind, we have $4 choose 3$ = $4$ ways

Step 3 - Choose a card from the deck that's not the same kind as in step (2): $52-4 choose 1$ = $48$ ways

Step 4 - Choose a card from the deck that's not the same kind as in step (2),(3): $52-4-4 choose 1$ = $44$ ways.

This comes to $109824$ but the actual answer is $54912$, so I'm overcounting by a factor of 2 somewhere.

Any help is appreciated!

You want to choose three suits of one kind, and a suit of each for two different kinds.

$$binom{13}1binom 43binom {12}2binom 41^2$$

The factor of two is due to the fact that the singletons are selected without order.
$$binom{13}1binom 43dfrac{(12cdot 4)~(11cdot 4)}{2!}$$

My intuition about it is that I pick any card, I match the other two, and then I make sure that from the remaining cards I don't make a full house or a four of a kind.

1. Pick any card. $binom{52}{1}$
   


2. Pick any matching 2 out of 3. $binom{3}{2}$
   


3. Pick a random card from the remaining 48 ones. $binom{48}{1}$
   


4. Pick a random card from the remaining 44 ones. $binom{44}{1}$
   

Now this solution is overcounting by 3! because the first card picked can be any of the 4 possible suites, which means it will repeat all those possible permutations.
So if you put all of these together:
$$binom{52}{1}binom{3}{2}binom{48}{1}binom{44}{1}frac{1}{3!}$$