# Lesson Explainer: Counting Principles: Addition Rule Mathematics

In this explainer, we will learn how to find the number of all possible outcomes of 2 or more events together using the addition counting principle.

Say that there are two restaurants in our neighborhood, Pizza Shop and Soup Kitchen. Pizza Shop has 10 different pizzas on its menu, and Soup Kitchen has 7 soups on its menu.

We want to find the number of different options for lunch if we buy our lunch from either of these restaurants. We can add the number of items in both restaurants’ menus to see that we have different options for lunch.

Above, we observed that we can count the number of distinct options for lunch from the two restaurants by finding the sum of distinct options from each restaurant. This is true because there is no common lunch item sold in both restaurants. Let’s put this in general terms.

Two events are said to be mutually exclusive if there is no common outcome from the two events. In context of our previous example, the two events are “buying a lunch item from Pizza Shop” and “buying a lunch item from Soup Kitchen.” Pizza Shop only sells pizzas, while Soup Kitchen only sells soups. Hence, the two events cannot produce a common outcome. This means that the events are mutually exclusive. For counting problems of two mutually exclusive events, we can use the addition rule.

### Theorem: Addition Rule for Two Events

Let and be mutually exclusive events. If there are distinct outcomes of event and outcomes of event , then there are distinct outcomes from either or .

Let us return to our example of buying a lunch item from either Pizza Shop or Soup Kitchen. There are distinct outcomes of buying a lunch item from Pizza Shop, whereas there are distinct outcomes of buying a lunch item from Soup Kitchen. Hence, the addition rule says that there are different outcomes.

In many counting problems, we can use the addition rule together with other counting rules. Let us recall the combination and permutation rules.

### Definition: Combinations and Permutations

A combination is the number of different ways to select objects from the total of distinct objects, where the order of the objects does not matter. This number is denoted by .

A permutation is the number of different ways to arrange objects in order from the total of distinct objects. This number is denoted by .

In our first example, we will apply the combination rule and the addition rule together.

### Example 1: Counting Outcomes of Two Events Using the Addition Rule

There are 10 boys and 6 girls. What is the numerical expression that allows us to calculate how many ways there are of forming a group that consists of either 3 boys or 2 girls?

We recall the addition rule for two events: if two events are mutually exclusive, then the number of distinct outcomes from either of the two events is given by the sum of the numbers of distinct outcomes from the two events.

In this example, the events are “forming a group consisting of 3 boys” and “forming a group consisting of 2 girls.” We note that there cannot be a common outcome of the two events, which tells us that they are mutually exclusive. Hence, we can apply the addition rule to obtain our answer if we know the number of outcomes from both events.

We need to count the number of outcomes from both events. We recall that the number of different ways to select objects from the total of distinct objects, where the order of the objects does not matter, is denoted by .

First, consider the event of forming a group consisting of 3 boys. Since we are choosing 3 boys from 10 boys, where the order of the boys does not matter, the number of different outcomes of this event is given by the combination .

Next, consider the event of forming a group consisting of 2 girls. Since we are choosing 2 girls from 6 girls, where the order of the girls does not matter, the number of different outcomes of this event is given by the combination .

Applying the addition rule, the number of ways to form a group that consists of 3 boys or 2 girls is

This is option B.

In the previous example, we applied the addition rule for two mutually exclusive events. We can generalize the addition rule to apply it to more than two events. To do this, we must first understand how the condition of mutual exclusivity can be extended when three or more events are involved. In this case, we need every pair of events to be mutually exclusive. In other words, three or more events are pairwise mutually exclusive when there is no outcome that is shared by any two events.

Let us return to our problem of buying a lunch. Previously, we assumed that our town only has two restaurants: Pizza Shop, which offers 10 different menu items, and Soup Kitchen, which offers 7 different menu items. Now, a new restaurant, Sandwich Place, has opened, with 5 different sandwiches on its menu. We can see that there is no lunch item that is shared by any two restaurants, so the events of buying lunch from different restaurants are pairwise mutually exclusive. Then, we have different options for lunch.

Likewise, we can extend the addition rule for pairwise mutually exclusive events.

The number of distinct outcomes from the collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event.

In the next example, we will apply the addition rule to three pairwise mutually exclusive events.

### Example 2: Counting Outcomes of Three Events Using the Addition Rule

What is the numerical expression we would use to find in how many ways can 4 balls of the same color be selected from 10 blue balls, 6 green balls, and 7 red balls? Assume none of the balls are identical.

We notice that there are 3 events that lead to selecting 4 balls of the same color: selecting 4 blue balls, selecting 4 green balls, and selecting 4 red balls. There is no outcome shared by two different events, so the three events are pairwise mutually exclusive.

We recall the addition rule that says that the number of distinct outcomes from the collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event.

So, if we know the number of distinct outcomes for each of the 3 events, we can obtain our answer by summing the three numbers.

Let us begin by finding the number of ways to select 4 blue balls. We recall that the number of different ways to select objects from the total of distinct objects, where the order of the objects does not matter, is denoted by . Since we are choosing 4 balls from 10 blue balls, the number of distinct outcomes of this event is .

Next, we consider the number of ways to select 4 green balls. Since we are choosing 4 balls from 6 green balls, the number of distinct outcomes of this event is .

Finally, we are selecting 4 balls from a total of 7 red balls, so the number of distinct outcomes of this event is .

By the addition rule, we can add these numbers to obtain the number of different ways to select 4 balls of the same color. Thus, we have

This is option D.

We can also use the addition rule with the fundamental counting principle, which is also known as the multiplication, or product rule. While the addition rule require the events to be mutually exclusive, the fundamental counting principle requires the events to be independent. We recall that two events are independent if a specific outcome of one event does not change the number of possible outcomes of the other event.

### Theorem: Fundamental Counting Principle

If we have two independent events and such that the number of possible outcomes for event is and the number of possible outcomes for event is , the total number of distinct possible outcomes of these two events together is the product .

Let us consider an example where we use the addition rule with the fundamental counting principle.

### Example 3: Counting Outcomes of Events Using the Addition Rule and the Fundamental Counting Principle

A cup contains 10 blue marbles, 6 green marbles, and 7 red marbles. None of the marbles in the cup are identical. How many ways can 4 marbles be chosen from the cup so that exactly 3 of them are the same color?

We note that there are three different events that lead to the described outcome:

• selecting 4 marbles so that exactly 3 of them are blue,
• selecting 4 marbles so that exactly 3 of them are green,
• selecting 4 marbles so that exactly 3 of them are red.

We can see that no outcome can be shared by two different events, so these events are pairwise mutually exclusive.

We recall the addition rule: the number of distinct outcomes from the collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event.

Let us count the number of outcomes in each event, starting with the event where we select exactly 3 blue marbles. This event can be split into two events: “selecting 3 blue marbles” and “selecting 1 marble that is not blue.” Since a specific outcome of one event does not affect the number of possible outcomes of the other event, these are independent events. By the fundamental counting principle, the number of outcomes of the two events together is the product of the number of outcomes from the two events.

Since there are marbles that are not blue, there are 13 different outcomes of selecting 1 marble that is not blue. To find the number of outcomes for selecting 3 blue marbles, we recall that the number of different ways to select objects from the total of distinct objects, where the order of the objects does not matter, is denoted by . Hence, there are ways to choose 3 marbles from 10 blue marbles. Applying the fundamental counting principle, the number of ways to select 4 marbles so that exactly 3 of them are blue is

We can count the number of outcomes from the other two events similarly. Next, we consider the number of ways to select 4 marbles so that exactly 3 of them are green. We split this event into two independent events: “selecting 3 green marbles” and “selecting 1 marble that is not green.” There are different ways to choose 3 marbles from 6 green marbles, and there are different marbles that are not green. By the fundamental counting principle, the number of ways to select 4 marbles so that exactly 3 of them are green is

Finally, we count the number of different ways to select 4 marbles so that exactly 3 of them are red. There are different ways to choose 3 marbles from 7 red marbles, and there are different marbles that are not red. By the fundamental counting principle, the number of ways to select 4 marbles so that exactly 3 of them are red is

Hence, applying the addition rule, the number of different ways 4 marbles can be chosen from the cup so that exactly 3 of them are the same color is

Let us consider another example where the addition rule and the fundamental counting principle are used together.

### Example 4: Counting Outcomes of Events Using the Addition Rule and the Fundamental Counting Principle

In how many ways can a group of 6 people be formed from 5 teachers and 10 parents, such that the group has at least one parent but less than four teachers?

We note that there are three different events that lead to the described outcome:

• forming a group with 1 teacher and 5 parents,
• forming a group with 2 teachers and 4 parents,
• forming a group with 3 teachers and 3 parents.

We can see that no outcome can be shared by two different events, so these events are pairwise mutually exclusive.

We recall the addition rule: the number of distinct outcomes from the collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event.

Let us count the number of outcomes in each event, starting with the event where we form a group with 3 teachers and 3 parents. This event can be split into two events: “forming a group with 3 teachers” and “forming a group with 3 parents.” Since a specific outcome of one event does not affect the number of possible outcomes of the other event, these are independent events. By the fundamental counting principle, the number of outcomes of the two events together is the product of the number of outcomes from the two events.

We recall that the number of different ways to select objects from the total of distinct objects, where the order of the objects does not matter, is denoted by . Hence, there are ways to form a group of 3 from 5 teachers, and ways to form a group of 3 from 10 parents. Applying the fundamental counting principle, the number of ways to form a group with 3 teachers and 3 parents is

We can count the number of outcomes from the other two events similarly. Next, we consider the number of ways to form a group with 2 teachers and 4 parents. There are different ways to form a group of 2 from 5 teachers, and different ways to form a group of 4 from 10 parents. By the fundamental counting principle, the number of ways to form a group with 2 teachers and 4 parents is

Finally, we count the number of different ways to form a group with 1 teacher and 5 parents. There are different ways to select 1 from 5 teachers, and different ways to form a group of 5 from 10 parents. By the fundamental counting principle, the number of ways to form a group of 1 teacher and 5 parents is

Hence, applying the addition rule, the number of different ways to form a group of 6 people from 5 teachers and 10 parents such that the group has at least one teacher but less than four teachers is

This is option E.

In previous examples, we have considered counting problems where we used combinations together with the addition rule. In our final example, we will consider a counting problem where we use permutations together with the addition rule and the fundamental counting principle.

### Example 5: Counting Outcomes of Events Using the Addition Rule and the Fundamental Counting Principle

Write the calculation we would use to work out the number of ways we can park 2 cars and then at least 2 trucks in 5 parking slots in a row.

We note that there are two different events that lead to the described outcome:

• parking 2 cars and 3 trucks,
• parking 2 cars and 2 trucks.

We can see that the two events do not share a common outcome, so these events are mutually exclusive.

We recall the addition rule for two events: if two events are mutually exclusive, then the number of outcomes from the two events is the sum of the number of distinct outcomes from each event.

Let us count the number of outcomes in each event, starting with the event where we park 2 cars and 3 trucks. This event can be split into two events: “parking 2 cars” and “parking 3 trucks.” A specific outcome of one event does not affect the number of possible outcomes of the other event. More specifically, if we park 2 cars in any two parking slots, it still leaves 3 slots for the trucks to park. Hence, these are independent events. The fundamental counting principle tells us that the number of outcomes of the two independent events together is the product of the number of outcomes from the two events.

We recall that the number of different ways to arrange objects in order from the total of distinct objects is denoted by , which is the permutation. Let us consider the problem of counting the number of different ways to park 2 cars in 5 parking slots in the context of permutation. For this purpose, we can label one of the cars “car 1” and the other car “car 2.” Then, the event of parking these two cars into two of the 5 available slots is equivalent to the event of choosing 2 out of 5 available parking slots and arranging them in order of “car 1” and “car 2.” Hence, the number of different ways to park 2 cars in 5 available parking slots is .

Having parked 2 cars, we have 3 parking slots available. Similar arguments as before give us the number of different ways to park 3 trucks in 3 available parking slots to be . Hence, by the fundamental counting principle, the number of different ways to park 2 cars and 3 trucks in 5 parking slots in a row is

Next, we consider the number of different ways to park 2 cars and 2 trucks. We know that there are different ways to park 2 cars in 5 available parking slots and different ways to park 2 trucks in the 3 remaining parking slots. Then, the fundamental counting principle tells us that the number of different ways to park 2 cars and 2 trucks in 5 parking slots in a row is

Finally, applying the addition rule, the number of different ways to park 2 cars and at least 2 trucks in 5 parking slots in a row is

This is option A.

Let us recap a few important concepts from this explainer.

### Key Points

• Let and be mutually exclusive events. If there are distinct outcomes of event and outcomes of event , then there are distinct outcomes from either or .
• The number of distinct outcomes from the collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event.
• The addition rule can be applied together with the fundamental counting principle to solve more complicated problems involving combinations and permutations.