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How many different ways can a person draw 5 cards from a standard deck of card without replacing them?



There are 52 * 51 * 50 * 49 * 48 different ways to select 5 cards (without replacement.)

The number 52 * 51 * 50 * 49 * 48 may also be represented as 52! / 47! using factorial notation:

52! / 47! = 52 * 51 * 50 * 49 * 48 * 47! = 52 * 51 * 50 * 49 * 48
47!




Now we ask a different question. How many different sets of 5 cards may be chosen from a deck of 52 cards?

One must observe that Ace of Spades King of Spades Queen of Spades Jack of Spades 10 of Spades
is the same hand as Queen of Spades 10 of Spades King of Spades Ace of Spades Jack of Spades


since they are the same cards, just rearranged in a different order.

In Permutations, we learned that 5 cards may be arranged 5! different ways. Thus, we must divide the number of ways to select 5 cards from a deck of 52 by 5! to get the number of five card hands that are possible:

52! / 47! 52! the number of 5 card hands
______ = ______ =
5! 47! * 5!


When one is selecting a set of items from a larger group (i.e. cards from a deck) and the order of the items does not matter, it is a combination.




Example Janice is holding the 7 cards below in her hand, and tells you to pick three of them. How many different ways can you do this?

4 of Clubs 8 of Diamonds 3 of Clubs Jack of Hearts 5 of Hearts King of Spades 9 of Spades


Solution There are 7 ways to select the first card, 6 ways to select the second card, and 5 ways to select the third card. Therefore there are 7 * 6 * 5 = 7! / 4! ways to select three cards. However, some of these selections will be the same! For example, the cards below show all the different ways to choose a three of clubs, a king of spades, and an eight of diamonds:

3 of Clubs King of Spades 8 of Diamonds King of Spades 3 of Clubs 8 of Diamonds 8 of Diamonds King of Spades 3 of Clubs
3 of Clubs 8 of Diamonds King of Spades King of Spades 8 of Diamonds 3 of Clubs 8 of Diamonds 3 of Clubs King of Spades


Since this one hand has 3! different permutations, we must divide 7! / 4! by 3! to get the number of possible hands that can be selected from Janice.

7! / 4! 7! 35 different hands may be picked from Janice
______ = ______
3! 4! * 3!


In General If one has n distinct objects, each selection or combination of r of these objects where order does NOT matter corresponds to

C( n, r ) = n! / (n - r)! n!
______ = ______ =
n! r! * (n - r)!


Some things to note about the combinations formula:




Want to see if you have conquered combinations? Try out some Combinations Practice Problems.




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