*Contributor: Mason Smith. Lesson ID: 11312*

How does order FACTOR into probability? Solving permutations and combinations requires the use of factorials. Find out what this means and how it might help you buy cereal or string a bunch of beads!

categories

subject

Math

learning style

Visual

personality style

Lion

Grade Level

Middle School (6-8), High School (9-12)

Lesson Type

Quick Query

Q: What do you call a hairstyle gone bad?

A: A perm mutation!

*Permutations* and *combinations* allow us to predict the probability of different events and outcomes with regards to *order*, or with respect to which outcome comes first, second, etc.

Permutations and combinations allow us to determine how large our sample space, or set of possible results, is. For more on sample space, check out the **Related Lesson** on experimental probability at the beginning of this *Probability: An Overview* series.

**Permutations**

A permutation is a grouping of outcomes where the order of the outcomes does matter.

- For example, Karen is painting her bedroom. She has orange, green, and purple paint. She plans to use one color as a base coat and stencil a design with another. How many different ways can she do this?

If we list all the groupings into the steps Karen must take to achieve this effect, we would start with the base color first, then the stencil color, like this: orange-green, orange-purple, green-orange, green-purple, purple-orange, and purple-green, which would be our sample space. We can see that there are six different ways for Karen to paint her room.

Let's look at another example:

- Sarah is stringing three different types of beads on a bracelet. If Sarah is using three different types of beads, how many different combinations can she use when stringing her beads?

Since order is important, we consider this a permutation. If we give each type of bead a number, such as 1, 2, 3, then we can list out all the groupings: 123, 132, 213, 231, 312, 321, and we see that there are six different ways, or orders, in which she can string the beads.

We can use an easy formula to solve for the number of permutations, but before we can start using the formulas for permutations or combinations, we have to know what *factorials* are, because both formulas use factorials.

**Factorials **

When we solve for combinations and permutations, we have to take into account factorials, which are expressed with an exclamation symbol (!).

A factorial of a number is the product of that number, times (or multiplied by) all numbers less than it until one. For example 5! (pronounced *5 factorial*) is just math shorthand for 5*4*3*2*1, or 120.

Here is the formula for permutations, where *n* is the *total number* of things we are picking from, and *r *is the number we *picked*:

nPr |
= | n! |
= | total! |

n - r! |
total - picked! |

A little trick for using the formula: If we want to solve two factorials in a fraction, instead of multiplying them out and dividing, we can list them in their multiplication form like this:

5! | = | 5 * 4 * 3 * 2 * 1 | = | 5 * 4 * 3 * 2 * 1 | = | 5 * 4 * 3 | = | 60 |

2! | 2 * 1 | 2 * 1 |

Then if a number is on both the top and bottom of the fraction bar, we can cross it off and just multiply the remaining numbers together.

Let's do an example together:

- A team is picking a president, vice president, and a secretary from a list of 8 people. How many ways can the three officers be chosen?

Think first what am I doing. Since being president and vice president are not the same thing, order matters. So, this example will be a permutation. There are three people picked, or *r = *3, and eight total, *n* = 8.

First the formula:

8P3 |
= | 8! | = | 8! |

(8 - 3)! | 5! |

Then we'll solve:

8 * 7 * 6 * 5 * 4 *3 *2 * 1 | = | 8 * 7 * 6 | = | 336 |

5 * 4 * 3 * 2 *1 |

Let's look at another example. Try to solve before looking at my example and discuss your response with a teacher or parent:

- Lee brings 7 playlists numbered 1-7 to a party. How many different ways can he choose the first 4 playlists to play?

Think: 7 total choices, 4 being picked

First set up the formula:

7P4 |
= | 7! | = | 7! |

(7 - 4)! | 3! |

Then solve:

7 * 6 * 5 * 4 * 3 * 2 * 1 | = | 7 * 6 * 5 * 4 | = | 840 |

3 * 2 * 1 |

Now that we know how to solve for the number of outcomes *with* order using permutations, let's learn how to solve for the number of outcomes *without* order using combinations.

**Combinations**

A combination is a grouping of outcomes, where the order of the outcomes does not matter.

- For example, we have a street vendor who sells cashews, peanuts, and almonds. How many different ways are there to mix the nuts?

Well, he could have cashews and almonds, cashews and peanuts, peanuts and almonds, almonds and peanuts, peanuts and cashews, and almonds and cashews.

Since the order doesn't matter (they are still the same types of nuts mixed together), we can eliminate all the options that repeat in a different order, and we are left with just the options cashews and almonds, cashews and peanuts, and peanuts and almonds.

This means that there are three ways of combining two types of nuts.

Try the following example. The answer will follow:

- Nathan wants to order a sandwich with two of the following ingredients: mushroom (m), eggplant (e), tomato (t), and avocado (a). How many different sandwiches can Nathan choose?

6 (m&e, m&t, m&a, e&t, e&a, t&a)

Here is the formula for combinations, where *n* is the total number of things we are picking from and* r* is the number we picked:

nCr |
= | n! |
= | total! |

r!(n - r)! |
number we pick!number we don't! |

I prefer to think of the formula in terms of what we do and don't pick rather than remembering what all the variables stand for.

Let's do an example:

- A team will form a 3-person committee from a list of 8 people. How many ways can the team choose 3 people? The position does not matter, so this is a combination situation.

Think of the problem like this: There are 8 people, we are picking 3.

8C3 |
= | 8! | = | 8*7*6*5*4*3*2*1 | = | 8 * 7 * 6 | = | 336 | = | 56 |

3!5! | 3*2*1*5*4*3*2*1 | 3 * 2 *1 | 6 |

Look at this other example:

- How many different kinds of punch can be made from 2 of the following: cranberry, apple, orange, banana, and grape juice.

Think: there are 5 options, we are picking 2.

5C2 |
= | 5! | = | 5 * 4 * 3 * 2 * 1 | = | 5 *4 | = | 20 | = | 10 |

2!3! | 2 * 1 * 3 * 2 * 1 | 2 * 1 | 2 |

Now that we understand how to solve combinations in probability, let's try some practice.

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