Combinations, formula, practice & examples

Arrangements without Order

In math, a combination is an arrangement in which order does not matter. Often contrasted with permutations, which are ordered arrangements, a combination defines how many ways you could choose a group from a larger group.

Note: To understand this topic, it is highly advisable to be familiar with factorial notation .

The formula for combinations

To find all of the differennt ways to arrange r items out of n items. Use the combination formula below. (n stands for the total number of items; r stands for how many things you are choosing.)

Combination Formula

Applied Example: Five people are in a club and three are going to be in the 'planning committee,' to determine how many different ways this committee can be created we use our combination formula as follows:

Combination Worked Out

Point of Contrast: The committee is a common theme for combination problems because, often, it does not matter how your committee is arranged. In other words, it generally does not matter whether you think about your committee as Sue, John and Jimmy or as John, Sue and Jimmy. In either case, the same three people are in the committee. Contrast this with permutations and our example of a locker pass code. Unlike a committee, a pass code of a lock depends on the order which will dictate whether or not you can open up your lock!

Practice Problems: combinations

Problem 1

Directions: Apply the combination formula to solve the problems below.

In a class of 10 students, how many ways can a club of 4 students be arranged?

Combination Formula Applied
Problem 2

Eleven students put their names on slips of paper inside a box. Three names are going to be taken out. How many different ways can the three names be chosen?

Eleven Choose Three
Problem 3

Over the weekend, your family is going on vacation, and your mom is letting you bring your favorite video game console as well as five of your games. How many ways can you choose the five games if you have 12 games in all?

12C5 = (12)!/(5!×(12-5)!)=(12)!/(5!×(7)!)
(12×11×10×9×8×7!) /(5!×7!) = 792

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