In a standard deck of $52$ cards, there are $4$ suits: clubs, diamonds, hearts and spades. For each suit, there are $13$ numbers ranging from A (Ace), 2, 3, ..., 10, J (Jack), Q (Queen), K (King). In the game Bridge, each player gets $13$ cards from a standard deck of $52$ cards. How many sets of $13$ cards are there where

a. all $13$ cards have the same suit?
b. the $4$ aces are part of the $13$ cards?
c. none of the $13$ cards have the same number?
d. exactly seven of the $13$ cards are spades?
e. at least seven of the $13$ cards are spades?

I think I understand A, $C(52,13) \cdot 4$, and B would be $C(4,4) + C(48,9) + C(47,8) \ldots + C(40,1)$. I feel like C would be similar to A but I'm not sure how, and for D and E I am completely lost, I don't know when I should multiply combinations and when I should use permutations.

  • 3
    $\begingroup$ Isn't the answer to (a) just $4$? $\endgroup$ Commented Nov 10, 2018 at 20:42
  • $\begingroup$ A set of cards has no specific order. $\endgroup$
    – hardmath
    Commented Nov 11, 2018 at 6:28

1 Answer 1


a) There are only $4$ ways that all $13$ cards can have the same suit because there are only four possible suits (we can have all $13$ be hearts, spades, diamonds, or clubs).

b) If we know that all $4$ aces are part of the $13$ cards, then we are really only choosing $9$ cards out of $48$. We are choosing $4$ of the $4$ aces, and then there are $48$ remaining cards, from which we are choosing $9$. So, the answer is ${4\choose 4}{48\choose 9}$.

c) There are $13$ possible numbers. The first card we choose can take any number ($13$ options). The second card we choose can take any number except for that of the first number $(12$ options), and so on. The answer is $13!$ because there are $13 \cdot 12 \cdot 11 \cdot \cdots 1$ ways to do this.

d) There are $13$ total spades in the deck, and we want exactly $7$ of them. So, there are ${13 \choose 7}$ ways to select the spades. We still have six more cards that can be anything but spades, though. There are ${39 \choose 6}$ ways to select them. Therefore, our final answer is ${13 \choose 7} \cdot {39 \choose 6}.$

e) Now, we want at least seven of the cards to be spades. So, our answer becomes the number of ways to have exactly $7$ spades plus the number of ways to have $8$ spades, all the way up to $13$. Mathematically, this is given by ${13 \choose 7} \cdot {39 \choose 6} + {13 \choose 8} \cdot {39 \choose 5} + {13 \choose 9} \cdot {39 \choose 4} + {13 \choose 10} \cdot {39 \choose 3} + \ldots$.

  • $\begingroup$ In (e) the $38$s should be $39$s. $\endgroup$ Commented Nov 11, 2018 at 5:34
  • 2
    $\begingroup$ Your answer for part (c) is incorrect. The order in which the cards are selected does not matter. How many ways can you select one card with each rank? $\endgroup$ Commented Nov 11, 2018 at 23:17
  • 2
    $\begingroup$ Part c is $4×4×...×4=4^{13}$ $\endgroup$ Commented Aug 19, 2021 at 21:14
  • $\begingroup$ Don't search for too complex answers : all cards from same suit : it can be Club, or Diamond or Hearth or Spade. 4 sets. $\endgroup$
    – Lourrran
    Commented Feb 6, 2023 at 13:50

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