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My problem is counting the number of ways in which a pair can occur. The answer in the back says $${13 \choose 2}{4 \choose 2}{4 \choose 2}44.$$ I get $${13 \choose 2}{4 \choose 2}{4 \choose 2}48.$$

I don't understand where the 44 comes from? The 48 = 52-4 comes from not selecting the 4 cards that we obtained. If someone could just explain where the 44 comes from that would be great.

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  • $\begingroup$ It depends on your definition of "two pairs". Does the hand "55666" count as two pairs? Or is it a full house? $\endgroup$
    – Sambo
    Commented Oct 27, 2022 at 14:25
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    $\begingroup$ Would it be easier written ${13 \choose 2}{11\choose 1} \times {4 \choose 2}{4 \choose 2}{4 \choose 1}$? The $44$ is ${52-8 \choose 1}$ or ${13-2\choose 1}\times{4 \choose 1}$ to select the card whose value appears once after two values (i.e. eight cards) are not possible as these values have been selected for the two pairs $\endgroup$
    – Henry
    Commented Oct 27, 2022 at 14:30
  • $\begingroup$ $4$ of your $48$ remaining cards would lead to a full-house rather than two pairs $\endgroup$
    – Henry
    Commented Oct 27, 2022 at 14:35
  • $\begingroup$ Keep in mind that in standard Poker terminology, saying that a hand is of type X precludes all stronger types. Thus $XXXYY$ is not two pairs just as $XXXYZ$ is not one pair. $\endgroup$
    – lulu
    Commented Oct 27, 2022 at 14:42
  • $\begingroup$ Would it be easier written....? yes that makes perfect sense $\endgroup$
    – HMPtwo
    Commented Oct 27, 2022 at 20:19

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