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.
$\frac{\binom{13}{1} \times \binom{4}{3} \times \binom{12}{1} \times \binom{11}{1} \times \binom{4}{1} \times \binom{4}{1} }{\binom{52}{5}}$

It is not $$\frac {\binom{13}{1}\cdot\binom{4}{3}\cdot\binom{48}{1}\cdot\binom{44}{1}}{\binom{52}{5}}$$ because you are considering the order of the last two selected cards say $A$ and $B$. In other words, you are counting every pair $\{A,B\}$ twice: as $(A,B)$ and $(B,A)$. 
Just divide by two and you will obtain the correct result.
