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!} |
= | 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 | |

is the same hand as |

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

^{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

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?

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 =

Since this one hand has 3! different permutations, we must divide

^{7!} / _{4!}
| 7! | 35 different hands may be picked from Janice |

______ | = ______ | |

3! | 4! * 3! |

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:

- C(n,r) is read " n choose r."

The symbol ( n ) r is frequently used in place of C(n,r) - C(n,0) = 1. This makes sense, because if you have n objects, and you select zero of them, there is only 1 way to do it - don't select any of the objects!
- Similarly, C(n, n) = 1. If you have n objects, and you select all n of them, there is only one way to do it - grab all of them!

Want to see if you have conquered combinations? Try out some