<s>[INST] Count the number of one pair hands in a standard deck In an attempt to answer this question, I tried the following solution:


*

*Let $A$ denote the number of one-pair hands

*Let $B$ denote the number of two-pair hands

*Let $C$ denote the number of three-of-a-kind hands

*Let $D$ denote the number of full-house hands

*Let $E$ denote the number of four-of-a-kind hands

*Let $F$ denote the total number of hands


Then:


*

*$A=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{10}{1}\cdot\binom{4}{2}\cdot\binom{4}{1}\cdot\binom{4}{1}\cdot\binom{4}{1}=\color{red}{6589440}$

*$B=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot\binom{4}{1}=247104$

*$C=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{4}{3}\cdot\binom{4}{1}\cdot\binom{4}{1}=109824$

*$D=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{4}{3}\cdot\binom{4}{2}=3744$

*$E=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{4}{4}\cdot\binom{4}{1}=624$

*$F=\binom{52}{5}=\color{red}{2598960}$

The part marked red is obviously wrong, since the number of one-pair hands is larger than the total number of hands. I've been going around this in circles trying to find out where I went wrong, with no luck whatsoever. Now I'm starting to think that other parts are wrong as well.
Any idea where my mistake is?
Thanks
 [/INST] To choose a pair, you choose one of the thirteen kinds, choose two of the four suits from that kind, choose three of the remaining kinds, and choose one of the four suits from each of those kinds, which yields
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3 = 1,098,240$$
hands with one pair.
Say your hand is $\color{red}{4\heartsuit}, \color{red}{4\diamondsuit}, \color{red}{J\diamondsuit}, 7\clubsuit, 5\spadesuit$.  Your mistake was that you did not take into account that there are $3! = 6$ orders in which you could select $\color{red}{J\diamondsuit}, 7\clubsuit, 5\spadesuit$.  Since each such selection would leave you with the same hand, you must divide your answer by $6$.
</s>