Using the Addition Principle

The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The Addition Principle tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle there are 8 total options.

Using the Multiplication Principle

The Multiplication Principle applies when we are making more than one selection. Suppose we are choosing an appetizer, an entrée, and a dessert. If there are 2 appetizer options, 3 entrée options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram.

The possible choices are:

1. soup, chicken, cake
2. soup, chicken, pudding
3. soup, fish, cake
4. soup, fish, pudding
5. soup, steak, cake
6. soup, steak, pudding
7. salad, chicken, cake
8. salad, chicken, pudding
9. salad, fish, cake
10. salad, fish, pudding
11. salad, steak, cake
12. salad, steak, pudding

We can also find the total number of possible dinners by multiplying.

We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle.

The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a permutation.

Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle

To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall.

There are four options for the first place, so we write a 4 on the first line.

After the first place has been filled, there are three options for the second place so we write a 3 on the second line.

After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product.

There are 24 possible permutations of the paintings.

Finding the Number of Permutations of n Distinct Objects Using a Formula

For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of [latex]n[/latex] objects and we want to choose [latex]r[/latex] objects from the set in order, we write [latex]P\left(n,r\right)[/latex]. Another way to write this is [latex]{}_{n}{P}_{r}[/latex], a notation commonly seen on computers and calculators. To calculate [latex]P\left(n,r\right)[/latex], we begin by finding [latex]n![/latex], the number of ways to line up all [latex]n[/latex] objects. We then divide by [latex]\left(n-r\right)![/latex] to cancel out the [latex]\left(n-r\right)[/latex] items that we do not wish to line up.

Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is [latex]6\times 5\times 4=120[/latex]. Using factorials, we get the same result.

[latex]\dfrac{6!}{3!}=\dfrac{6\cdot 5\cdot 4\cdot 3!}{3!}=6\cdot 5\cdot 4=120[/latex]

There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.

[latex]P\left(n,r\right)=\dfrac{n!}{\left(n-r\right)!}[/latex]

Note that the formula stills works if we are choosing all [latex]n[/latex] objects and placing them in order. In that case we would be dividing by [latex]\left(n-n\right)![/latex] or [latex]0![/latex], which we said earlier is equal to 1. So the number of permutations of [latex]n[/latex] objects taken [latex]n[/latex] at a time is [latex]\frac{n!}{1}[/latex] or just [latex]n!\text{.}[/latex]

Finding the Number of Permutations of n Non-Distinct Objects

We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be [latex]12![/latex] ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the [latex]12![/latex] permutations we counted are duplicates. The general formula for this situation is as follows.

[latex]\dfrac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}[/latex]

In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are [latex]4![/latex] ways to order the stars and [latex]3![/latex] ways to order the moon.

[latex]\dfrac{12!}{4!3!}=3\text{,}326\text{,}400[/latex]

There are 3,326,400 ways to order the sheet of stickers.

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