<s>[INST] Point out my fallacy in the three-of-kind problem, please How many hands of three-of-kind are there from a standard 52-card deck?
$$
13 \binom{4}{3} \binom{12}{2}\binom{4}{1}^2
$$
is the correct answer:


*

*$13$ :  #ways to choose the denomination of the three-of-a-a-kind

*$\binom{4}{3}$ :  #ways to choose their suits (three-of-a-a-kind);

*$\binom{12}{2}$ :  #ways to choose two distinct denominations for the remaining cards

*$\binom{4}{1}^2$ :  #ways to choose the suits of these latter two.


One could argue that the following is also correct
$$
52 \binom{3}{2} \binom{12}{2} \binom{4}{1}^2
$$
where


*

*$52$ :  #ways to choose both denomination and suit of the first card in the three-of-a-kind 

*$\binom{3}{2}$ :  #ways to choose the suit of the remaining two cards (in the three-of-a-kind)

*$\binom{12}{2} \binom{4}{1}^2$   as before


But 
$$
13 \binom{4}{3} < 52 \binom{3}{2} 
$$
This means that there is an overcounting in the #ways to select the three-of-a-kind.
Can anyone kindly indicate me what I am overcounting? Thank you.
 [/INST] The $52 {3 \choose 2}$ counts each three of a kind three times.  It counts each three of a kind once for having each of the three cards first.  And in fact $52 {3 \choose 2}=3 \cdot \left(13 {4 \choose 3}\right)$
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