A poker set consists of four cards, a deck of 52 cards is used.
My reasoning is:
First, we choose the rank which will appear twice in our hand. We can do it in $13$ ways. Now, we choose two out of four cards of the chosen rank to put in our deck, we can do it in $4\choose2$ ways.
What we are left with is to arrange the remaining cards so that they do not repeat: $48$ possibilities for the third card, $44$ possible cards for the fourth one and $40$ for the last one. Since the order does not matter, we divide by $3!$, so my answer is $$\frac{13 \cdot {4\choose2}\cdot48\cdot44\cdot40}{3!}$$
However, I'd like to know if I got this correct answer by sheer coincidence or my model is actually correct.

  • $\begingroup$ ILoveChess Yes I saw that, and have erased my comment. Your method works, in my opinion. +1 on question. $\endgroup$ – coffeemath Mar 25 '17 at 0:52
  • 2
    $\begingroup$ Your model is correct and can be explained by the fact that each arrangement is counted exactly six times in the number $13\cdot \binom{4}{2}\cdot 48\cdot 44\cdot 40$. The more complicated statement of the multiplication principle of counting allows for each outcome to be counted more than once (instead of exactly once) so long as each outcome is counted the same number of times at which point you divide by the number of times each was counted to get the correct total number of outcomes. $\endgroup$ – JMoravitz Mar 25 '17 at 0:55
  • $\begingroup$ Wikipedia gives the same number, calculated as ${13\choose1}{4\choose2}{12\choose3}{4\choose1}^3$. $\endgroup$ – Χpẘ Mar 25 '17 at 1:32
  • $\begingroup$ This is also confirmed in math.stackexchange.com/questions/567420/… and its answer. $\endgroup$ – David K Mar 25 '17 at 4:03

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