Poker, number of three of a kind, multiple formulaes

Question

I wanted to calculate some poker hands, for a three of a kind I infered, 1) every card rank can form a 'three of a kind' and there are 13 card ranks, 2) there are $\binom{4}{3}$ ways to choose three cards out of the four suits of every card rank, and 3) for the remaining card I can choose two out of 49 cards, i.e. $\binom{49}{2}$. Together the formulae is $$ 13 \cdot \binom{4}{3} \cdot \binom{49}{2} = 61152 $$ But on Wikipedia I found a different formulae, namely $$ \binom{13}{1} \binom{4}{3} \binom{12}{2} \left( \binom{4}{1} \right)^2 = 54912 $$ which makes also totally sense to me (1. card rank, 2. subset of suits, 3. choose form the left card ranks, 4. assign suits). But I can't see why my first formulae is wrong, can anybody explain this to me?

Answer 1

In counting the number of hands with three of a kind we must not include those that have four of a kind or a full house.

As @Jean-Sébastien notes in the comments, your formula counts $$\# (\textrm{three of a kind}) + 4\# (\textrm{four of a kind}) + \# (\textrm{full house})$$ or $$54,912 + 4\times 624 + 3,744.$$ The factor of four arises since $\rm AAA{\underline A} = AA{\underline A}A = A{\underline A}AA = {\underline A}AAA$.

Answer 2

We can count more or less like you did, using $\dbinom{13}{1}\dbinom{4}{3}\dbinom{48}{2}$ (note the small change), and then subtracting the full houses.

Or else after we have picked the kind we have $3$ of, and the actual cards, we can pick the two "useless" cards. The kinds of these can be chosen in $\dbinom{12}{2}$ ways. Once the kinds have been chosen, the actual cards can be chosen in $\dbinom{4}{1}^2$ ways, for a total of $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2.$$