# Number of $13$-card hands satisfying the stated conditions

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.

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

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 $${13 \choose 7} \cdot {39 \choose 6} + {13 \choose 8} \cdot {39 \choose 5} + {13 \choose 9} \cdot {39 \choose 4} + {13 \choose 10} \cdot {39 \choose 3} + \ldots$$.

• In (e) the $38$s should be $39$s. Commented Nov 11, 2018 at 5:34
• 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? Commented Nov 11, 2018 at 23:17
• Part c is $4×4×...×4=4^{13}$ Commented Aug 19, 2021 at 21:14
• Don't search for too complex answers : all cards from same suit : it can be Club, or Diamond or Hearth or Spade. 4 sets. Commented Feb 6, 2023 at 13:50