# 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 $4$? – Lord Shark the Unknown Nov 10 '18 at 20:42
• A set of cards has no specific order. – hardmath Nov 11 '18 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 $${13 \choose 7} \cdot {38 \choose 6} + {13 \choose 8} \cdot {38 \choose 5} + {13 \choose 9} \cdot {38 \choose 4} + {13 \choose 10} \cdot {38 \choose 3} + \ldots$$.
• In (e) the $38$s should be $39$s. – Lord Shark the Unknown Nov 11 '18 at 5:34