"How many ways are there to choose four cards of different suits and different values from a deck of 52 cards ? "

I solved this question in this way :- "So, for the first card we have 52 options to choose from .for eg. king of diamonds . so for the next card we can't choose from the suit of diamonds with the value of a king . Hence, there are 36 options to choose the second card . [ 52-13(suit of diamond)=39-3(rest of the kings)]. for eg. queen of hearts has been chosen as second. So, for choosing third card we have 21 options to choose from . [ 36-13(suit of hearts)=23-2(rest of the queens)] . for eg. ace of spades has been chosen as third . Now, for the last card we have 7 options to choose from . [ 21-13(suit of spades)=8-1(rest of the ace)] . Hence, the total no. of ways should be :- 52x36x21x7 = 275184 ways " I was told that this logic is wrong and there will be some overcounting . But, I couldn't understand why . So, can someone please point out the flaw in this method .

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    $\begingroup$ Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. $\endgroup$ Aug 19, 2023 at 12:35
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    $\begingroup$ For example, if you choose the king of diamonds on the first draw and then three other cards, that is the same as if you chose one of the other cards on the first draw and the king of diamonds later. $\endgroup$
    – John Douma
    Aug 19, 2023 at 12:47
  • $\begingroup$ $$\binom{13}{4} \times 4!$$ The first factor reflects the number of different ways of selecting the $~4~$ ranks from $~13~$ ranks, sampling without replacement, where order of selection is deemed irrelevant. The second factor reflects the number of ways of ordering the distribution of the four suits among the four selected ranks. $\endgroup$ Aug 19, 2023 at 14:07

1 Answer 1


If we want to do it your way, two figures need correction, and you need to divide by $4!$ because the cards could have been chosen in various orders,

thus $\;(52\times 36\times22\times10)\div 4!$

As you can see, the approach you adopted is complex and error prone, and a much simpler way is to simply choose values of cards suit by suit,

thus $\;13\times12\times11\times10$


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