If your poker hands have five cards, we can count them as follows:
- Select the rank of the 3-of-a-kind: there are $\binom{13}{1}$ ways of doing this;
- Select the suits of the 3-of-a-kind: there are $\binom{4}{3}$ ways of doing this;
- Select two ranks from the remaining twelve ranks; there are $\binom{12}{2}$ ways of doing this;
- Select the suit of the higher of the two leftover cards; there are $\binom{4}{1}$ ways of doing this;
- Select the suit of the lower rank of the two leftover cards; there are $\binom{4}{1}$ ways of doing this.
This gives
$$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1} = 54912.$$
Your count is off among other reasons because you are considering the order in which you pick the remaining two cards (you should have $49\times 48/2$ in the first summand instead). But the real problem is that you are counting each four-of-a-kind hand four times. For example, if you have four aces and a king, you count it once when your three-of-a-kind are the aces of hearts, diamonds, and spades; then again when they are the aces of hearts, diamonds, and clubs; then again when they are the aces of hearts, spades, and clubs; and yet again when they are the aces of diamonds, spades, and clubs. If you subtract four times the number of 4-of-a-kind hands you get the correct answer.
\text
command. For instance,13\cdot48-\text{Full Houses}
produces "$13\cdot48-\text{Full Houses}$". $\endgroup$