<s>[INST] What is the probability that a 5-card poker hand contains 3 of a kind? I tried searching on the net and it shows an answer of
$\frac{ \binom{13}{1} \times \binom{4}{3} \times \binom{12}{2} \times \binom{4}{1} \times \binom{4}{1} }{ \binom{52}{5} }$
Why is it not the same as:
$\frac{ \binom{13}{1} \times \binom{4}{3} \times \binom{48}{1} \times \binom{44}{1}}{ \binom{52}{5} }$
This answer seems to be doubled of the correct answer based on the net. 
I also tried different approach and it resulted as this one which is also doubled of the correct answer based on the net.
$\frac{\binom{13}{1} \times \binom{4}{3} \times \binom{12}{1} \times \binom{11}{1} \times \binom{4}{1} \times \binom{4}{1} }{\binom{52}{5}}$
 [/INST] Let's say our 3-of-a-kind is made of 5s. When you say " $\binom{48}{1} \times \binom{44}{1}$ " , what you mean is "pick a card that isn't a $5$" (say, a $6$), and then "pick a card that isn't a $5$ or $6$".
The problem is that this counts every pair twice: if your final cards are $(4,7)$ then you count $(7,4)$ a separate time.
To get around this, select two cards $\binom{12}{2}$ and two suits $\binom{4}{1}^2$.
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