Number of 3-of-a-kind hands that are not 4-of-a-kind and not a full-house?

Question

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!

Answer 1

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

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

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

Answer 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: $$\binom{52}{1}\binom{3}{2}\binom{48}{1}\binom{44}{1}\frac{1}{3!}$$