Anonymous asked in Science & MathematicsMathematics · 5 years ago

Find the number of distinguishable subsets of a 52 card twenty one deck?

Twenty-one subsets. Find the number of distinguishable subsets of a

52-card twenty-one deck. (If the cards were distinguishable, as in poker for

example, the answer would be 2^52. But in twenty-one, suits are disregarded

and denominations 10, J, Q, and K are indistinguishable.)

The answer should be 33,203,125.

Have no idea how to reach this answer. Thank you!!

4 Answers

  • Anonymous
    5 years ago
    Favorite Answer

    What I'm reading is that you have 52 cards and classify them as

    4 each of Ace through Nine, that's 36

    16 cards worth 10, that's 16

    Total is 52

    I think you have to break it up into a lot of cases, depending on the size of the subset.

    0: 1 subset, the empty set.

    1: 10 different cards.

    2: 55, since you have 10 sets where the two cards are the same plus 10C2 = 45 sets where the two cards are different.

    3: 220. You have 10C3 - 10x9x8/6 = 120 sets where the 3 cards are different.

    You have 2 x 10C2 sets where 2 cards are the same and the 3rd is different--10C2 ways to pick two different cards X 2 ways to pick which one is doubled. 2x45 = 90

    You have 10 sets where all 3 cards are the same.

    120 + 90+ 10 = 220

    To continue through all cases will be a lot of work up to a point, then it will get easier for large subsets:

    52: 1 set

    51: 10 sets, since you have 10 choices for which card is missing.

    The case of 30 cards, e.g., will be quite complex, because you can have all of the 10 cards, or none. You have to count each possible number. And you have to look at how dupes of the remaining cards are split up.

    You may have opportunities to use some recursive logic. E.g, from 10 1-card subsets, you can add the same card in 10 ways, or a different card in 9 ways, giving you 10x9=90, but that double counts things so you divide by 2 to get 45. You get 10 + 45 = 55 as above.'

    You can probably do a similar breakdown to move from 55 to 220.

    There may be a different way to do this, but I suspect it will take some sort of advanced formula. There are a couple of combinatorics specialists around here. Maybe they will have an easier answer.

    This seems like quite a difficult problem. Where did you get it?

    • Lv 7
      5 years agoReport

      Yes, Ben really nailed it with a different, very elegant answer. It takes a bit of brilliance to find the really simple solutions, sometimes.

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  • Ben
    Lv 6
    5 years ago

    (To reverse engineer the solution it may be helpful to factor the answer.)

    Let's pretend that the ranks 10-K are distinguishable first. Then each subset is defined by how many A, how many 2, ..., how many K there are. There are 5 choices for each (0,1,2,3,4), so the total number of distinguishable subsets is 5^(13).

    Now take into account the indistinguishability of the upper ranks. Now they behave like a single larger rank, and a subset consists of any number from 0 to 16 of them. So the number desired is 5^(9)*17.

  • Nick
    Lv 6
    5 years ago

    All we care about is how many of each denomination there are in any given subset.

    There are 0,1,2,3 or 4 aces, 0,1,2,3 or 4 twos ... 0,1,2,3 or 4 nines and 0,1,2,..,16 ten value cards. Hence:

    5^9*17 = 33203125 <----

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  • Anonymous
    5 years ago


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