A standard​ 52-card deck of cards is shuffled and then a hand of 10 cards is dealt from the deck.

How many different outcomes are possible that have exactly $2$ ​pairs, and all other cards of different​ ranks?

Note that a pair consists of two cards of the same​ rank, and each pair must be of a different rank. An example of a​ 5-card hand with 2 pairs​ is: 7 of​ hearts, 7 of​ clubs, 9 of​ diamonds, 9 of​ clubs, and Queen of spades.

It seems we cannot choose two of the same pair first i start by choosing the two numbers that i will hold pairs of then discount the 8 cards from selection. Then i have 44 left 40 etc. i think i need to multiply by one over 6! to account for the ordering of the numbers i have i selected ?

$${{13}\choose {2}}\cdot44\cdot 40\cdot 36\cdot 32\cdot 28\cdot 24 \cdot \frac{1}{6!} = 147603456$$

I guess my confusion is i could draw the same 10 card hand say say i draw 5 6 7 8 9 3 if i drew 3 5 6 7 8 9 its the same 10 card hand in the end how do i fix that?

  • 1
    $\begingroup$ It may often help to first write out what you are looking to count. $~$ We favor a selection of two distinct pairs and six distinct singletons. $~$ That is: 2 from 4 suits for each of 2 from 13 kinds, and 1 from 4 suits for each of 6 from the 11 other kinds.$$\binom 42^2\binom {13}2\binom 41^6 \binom{11}6$$ $\endgroup$ – Graham Kemp May 30 '18 at 0:58

You forgot to choose which pairs of cards exactly you take from the two ranks first selected: factor ${4 \choose 2}{4 \choose 2}$.

Otherwise it is perfect.


One correct approach is as follows. We can uniquely construct any hand which satisfies the criteria by following the steps below in sequence:

  • Select 2 ranks, one for each pair; $\binom {13}2$ possibilities
  • Select suits for each pair; $\binom 42$ possibilities for each pair
  • Select ranks for the singleton cards; $\binom{11}{6}$ possibilities
  • Select suits for each lone card; $4$ possibilities for each card.

With that in mind, the total number of such hands will be $$ \binom {13}2 \cdot \binom 42^2 \cdot \binom {11}6 \cdot 4^6 = 5\,313\,724\,416 $$

  • $\begingroup$ Shouldn't that be $4^6$ ? @Omnomnomnom $\endgroup$ – Faust May 30 '18 at 0:21
  • $\begingroup$ And $\binom{11}{6}$ $\endgroup$ – aschepler May 30 '18 at 0:22
  • $\begingroup$ Good catch; just fixed it $\endgroup$ – Omnomnomnom May 30 '18 at 0:22
  • $\begingroup$ This works, but the OP's method is more efficient imho. (+1 though) $\endgroup$ – Arnaud Mortier May 30 '18 at 0:23
  • $\begingroup$ NVM i just cant use a calculator thanks all. $\endgroup$ – Faust May 30 '18 at 0:24

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