probability 5 card hand 
In a 5 card hand of cards, what is the probability that you get 3 of a kind? (Note: the other two cards have to be diferent from the 3 of a kind and each other).

Since there are 13 possible kinds (A, 2 to 10, K,Q,J), I did $13 * \binom{4}{3}*\binom{48}{1}*\binom{44}{1}/\binom{52}{5}$. Or I tried  $13 * \binom{4}{3}*\binom{12}{1}*\binom{4}{1}*\binom{11}{1}*\binom{4}{1}/\binom{52}{5}$However, the answer is approximately 0.0211, and I don't know why my answer (or rather, my way of thinking) is wrong. 
 A: As mentioned in the comments you're double counting some configurations. Hence you need to divide by $2! = 2$. Another way to notice that is to see that the probability you get with your method is twice the probability given in the solution.
A: To find the number of three of a kind hand, choose one of the $13$ ranks, choose three of the four cards of that rank, choose two of the remaining $12$ ranks, and choose a suit for each of those ranks.
$$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$
Hence, the probability of obtaining three of a kind when five cards are drawn is 
$$\frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}^2}{\dbinom{52}{5}}$$
In your calculation, you first chose one single card, then chose the other, so you made an ordered selection of the two single cards.  It does matter whether you choose $7\color{red}{\heartsuit}$ then $5\spadesuit$ or $5\spadesuit$ then $7\color{red}{\heartsuit}$.  Therefore, you must divide by the $2!$ orders in which you could have selected the single cards.  Notice that 
$$\frac{1}{2!}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1} = \binom{12}{2}\binom{4}{1}^2$$
