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Probability of Deck of Cards, ignoring suits

  1. Nov 23, 2014 #1
    I am looking for a way of calculating the possible combinations of a standard deck of 52 cards.

    I am aware of the 52! number, which is the total no of combinations 52 cards can form in a deck, but would like to know how to determine the total no of combinations if the suits are ignored.

    Thanks in advance.
     
  2. jcsd
  3. Nov 23, 2014 #2

    Doug Huffman

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    I do not see an expression for suits in 52!
     
  4. Nov 23, 2014 #3

    Stephen Tashi

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    Do you mean "permutations" instead of "combinations"?
     
  5. Nov 23, 2014 #4
    ??
    isnt the number of ways of arranging 52-card deck 52! ?
     
  6. Nov 23, 2014 #5

    Doug Huffman

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    The cards could as well be of 52 unique words, still with no suits.
     
  7. Nov 23, 2014 #6
    Yes, in fact I do.

    All I need is the way of calculating the number of ways of arranging the deck of cards without suits, i.e. Qh As 7d would be same as Qs Ad 7c.
     
  8. Nov 23, 2014 #7
    Sure, it would be the same problem, although we need to have 13 unique words, 4 each in such a deck of 52 cards to arrive to my initial question again.
     
  9. Nov 23, 2014 #8

    mathman

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    Wild guess: [itex]\frac{52!}{4!^{13}}[/itex]
     
  10. Nov 23, 2014 #9
    mathman,

    thanks and it looks like the right approach but would appreciate a little explanation, please.
    As an example, how would that formula be transformed for the deck of 8 cards with 4 aces and 4 kings?

    would that be 8!/4!^2?
     
    Last edited: Nov 23, 2014
  11. Nov 24, 2014 #10

    mathman

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    Yes. Numerator is all permutations of deck. Each term of denominator is permutation of cards of the same rank.
     
  12. Nov 24, 2014 #11
    I have manually double-checked the result from the formula for the deck of 8 and it is just spot on!

    Appreciate your help, thanks a lot!
     
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