# Intro to combinatorics, part 2: Combinations

Now that we’ve discussed the permutation, let’s meet its less-picky sibling, the *combination*.

Combinations are like permutations with one small difference: order does not matter. It’s like reaching into a bag of Skittles and picking out a handful: whether those pieces were the first you touched or the last, they’re all going to end up in your mouth.

Let’s pretend I’m running a lottery. I put 12 balls (lettered A through L) into a bag and pick 3 out at random. How many different ways of picking them are possible?

If the order mattered, as it would in a permutation, there would be

_{12}P_{3} |
= | 12! | = | 12! | = | 12 x 11 x 10 x |

(12-3)! | 9! |

or 1,320 possible permutations. (Note this is the same number as my horse-racing example from the previous post.)

But in this lottery, it doesn’t matter which ball comes out first, second, or third. Let’s say you buy a ticket with the letters D, F, and K. In how many different ways can those three balls come out to make you a winner?

D-F-K | F-D-K | K-D-F |

D-K-F | F-K-D | K-F-D |

There are 6 different* permutations* (ordered outcomes) involving the *combination *of D, F, and K. So, in a way, you are getting six permutations for the price of one. But then again, so is everyone else who’s playing the game.

## Eliminating duplicates

Because some permutations are effectively the same as others in our lottery, our 1,320 number is too high. We have 6 times as many combinations as we originally counted. Six, you might notice, is equal to 3! – or *k*! in this example.

The number of ways you can choose a *combination* of *k* objects from a set of *n* objects is equal to:

_{n}C_{k} |
= | n! |

k! x (n–k)! |

The formula is almost the same as that for a permutation; we’re just including the part in yellow, which makes us account for the “identical” orders.

Therefore, in my lottery, there are

_{12C}_{3} |
= | 12! | = | 12! | = | 12 x 11 x 10 x |

3! x (12-3)! | 3! x 9! | 6 x |

or 220 different possible *combinations*.

## English is stupid

There’s an easy way to remember the distinction between permutations and combinations. What is this thing called?

We call it a *combination* lock. But since you have to give three consecutive numbers in order, it should really be called a *permutation* lock. Funny, huh?

## Practice questions

Now, it’s time to see how well you’ve grasped this concept.

**Question 1.** I have 7 friends who need a ride, but I have room enough for only 4 in my car. (My friends don’t care where they sit, just that they get a spot.) How many different *combinations* of friends can I bring with me?

Correct response: | Show |
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**Question 2.** In a *Jeopardy!* tournament, the 10 non-winning quarterfinalists are eligible for four wild-card slots. It doesn’t matter whether you’re the #1 wild card or the #4 wild card, you’ll get the same reward: a trip to the semifinals. Therefore, order does not matter.

How many different *combinations* of wild cards can be made from the set of non-winners? (Another way to think about this: if I put the non-winners’ names into a hat and pulled out four, how many different groups of four could I possibly get?)

Correct response: | Show |
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In the next post: how I use combinatorics, probability, and a wealth of data from the J! Archive to derive optimal wagers in Jeopardy! tournaments.