Lesson Video: Counting Principles: Addition Rule | Nagwa Lesson Video: Counting Principles: Addition Rule | Nagwa

# Lesson Video: Counting Principles: Addition Rule Mathematics

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

16:40

### Video Transcript

In this video, weโll learn how to find the number of all possible outcomes of two or more events together using the addition counting principle.

Imagine that there are two restaurants in the neighborhood, Pizza Shop and Soup Kitchen, completely original names, of course. Pizza Shop has 10 different pizzas on its menu, whilst Soup Kitchen has seven soups on its menu. We want to find the number of different options for lunch if we choose to buy that lunch from either of these restaurants. And to do so, we simply add the number of items on both restaurantsโ menus. Thatโs 10 plus seven, which is equal to 17 different options for lunch.

Now, of course, this is only true because thereโs no common lunch item thatโs sold in both restaurants. If the Soup Kitchen, for instance, decided that they were going to suddenly sell pepperoni pizza, weโd need to rethink our calculations somewhat.

Letโs put this in general terms. Remember, a pair of events is said to be mutually exclusive if thereโs no common outcome from the two events. In the context of our previous example, those events would be buying a lunch item from the Pizza Shop and buying an item from the Soup Kitchen. If the Soup Kitchen, as we said, decided to sell pepperoni pizza, then these two events would not be mutually exclusive. As it was, since each event does not produce a common outcome, they are mutually exclusive.

So if a pair of events ๐ด and ๐ต are mutually exclusive and there are ๐ distinct outcomes of event ๐ด and ๐ distinct outcomes of event ๐ต, the total number of outcomes is ๐ plus ๐.

Now weโre going to want to apply this alongside the rules we have for combinations and permutations. A combination is found when weโre selecting objects from a large group and order doesnโt matter. So say we had three objects ๐ด, ๐ต, and ๐ถ. Choosing ๐ด then ๐ต would be the same as choosing ๐ต then ๐ด. And to find the number of ways of choosing ๐ items from ๐ when order doesnโt matter, we find ๐C๐. When order does matter, itโs a permutation. And the number of ways of choosing ๐ items from ๐ in this case is ๐P๐.

So with all this in mind, letโs apply the combination rule and the addition rule together.

There are 10 boys and six girls. What is the numerical expression that allows us to calculate how many ways there are of forming a group that consists of either three boys or two girls? Is it (A) 10 ๐ถ three times six ๐ถ two? Is it (B) 10 ๐ถ three plus six ๐ถ two? Is it (C) 10 ๐ three times six ๐ two, (D) 10 ๐ three plus six ๐ two, or (E) 10 ๐ถ three minus six ๐ถ two?

Weโre forming a group that consists of either three boys or two girls. So in fact, there are two events. The first event, letโs call that event ๐ด, is the event that we choose three boys from a total of 10. The second event, letโs call that ๐ต, is the event that we choose two girls from a total of six. There cannot be a common outcome of the two events. So they must be mutually exclusive. This tells us that weโre going to be able to apply the addition rule to answer this problem.

This says that if two events ๐ด and ๐ต are mutually exclusive, where ๐ด has ๐ distinct outcomes and ๐ต has ๐ distinct outcomes, the total number of outcomes from either of the two events is given by ๐ plus ๐. We simply add together the numbers of distinct outcomes from the two events. So our job now is to calculate the number of outcomes from each event.

Weโre choosing three boys from a total of 10. Now thereโs no indication here that order matters. In fact, letโs say we have boy one, boy two, and boy three. Switching the order in which we choose the first two boys so that we choose boy two then boy one then boy three doesnโt actually matter. We still end up with the same final three boys. And so weโre dealing with combinations. The number of combinations there are, which is when order doesnโt matter, of choosing ๐ items from ๐ is ๐C๐ or ๐ choose ๐.

And the notation that weโve used here will not necessarily be the notation youโre used to. Depending on where you are in the world, you might see both the ๐ and the ๐ as subscript or alternatively as an ordered pair and sometimes even as a column vector. So with this in mind, weโll calculate the number of ways of choosing three boys from a total of 10. Itโs 10 choose three. Similarly, the order in which the girls are chosen doesnโt matter. And weโre choosing two from a total of six. So itโs six choose two.

Since the events are mutually exclusive, we add these together to find the total number of possible outcomes. And so the number of ways of forming a group consisting of either three boys or two girls is 10 choose three plus six choose two, which is option (B).

We have now demonstrated how to apply the addition rule for a pair of mutually exclusive events. Letโs now see how we can generalize this to apply it to more than two.

Returning to our problem where weโre trying to decide what lunch to buy. Imagine now that a third restaurant opens, Sandwich Place. And this restaurant has five different types of sandwiches. The three events, which is choosing a pizza or choosing a soup or choosing a sandwich, are still mutually exclusive. There are no common items on these menus. The total number of options is now the sum of 10 and seven and five. There are 22 different options for lunch.

And so we can essentially extend our addition rule for pairwise mutually exclusive events. That is, the number of distinct outcomes from a collection of pairwise, mutually exclusive events is the sum of the number of distinct outcomes from each event. Letโs demonstrate this in our next example.

What is the numerical expression we would use to find in how many ways can four balls of the same color be selected from 10 blue balls, six green balls, and seven red balls. Assume none of the balls are identical. (A) 10 ๐ถ four times six ๐ถ four times seven ๐ถ four, (B) 10 ๐ four times six ๐ four times seven ๐ four. Is it (C) 10 ๐ four plus six ๐ four plus seven ๐ four, (D) 10 ๐ถ four plus six ๐ถ four plus seven ๐ถ four, or (E) 10 ๐ถ four times six ๐ถ four plus seven ๐ถ four?

Weโre selecting four balls from 10 blue, six green, and seven red. Now the key aspect of this question, which will help us answer it, is that none of the balls are going to be identical. So when we choose four balls, weโre either going to choose four blue, four green, or four red. Since no outcome is shared by the event of choosing four blue balls, four green balls, and so on, the three events are said to be pairwise mutually exclusive.

The addition rule says that the number of distinct outcomes from this collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event. So we need to work out the number of ways of choosing four blue balls from a total of 10, four green balls from a total of six, and four red balls from a total of seven. And then weโll add these values together.

Now since the order in which these balls are chosen doesnโt matter โ choosing, say, four blue balls will result in the same final outcome regardless of the order in which these are chosen โ we know that we have combinations. In other words, the number of ways of choosing ๐ items from a total of ๐ distinct items when order doesnโt matter is ๐ choose ๐. So the number of ways of choosing the four blue balls from a total of 10 is 10 choose four. The number of ways of choosing four green balls from a total of six is six choose four. And choosing four red balls from a total of seven is seven choose four.

The addition rule says that the total number of outcomes is the sum of these. So the numerical expression we will use is 10 ๐ถ four plus six ๐ถ four plus seven ๐ถ four. And thatโs option (D).

Now a powerful property of the addition rule is that it can be used alongside the fundamental counting principle. Thatโs sometimes also called the multiplication rule or the productโs rule for counting. Whilst the addition rule, weโve seen, requires the events to be mutually exclusive, the fundamental counting principle requires the events to be independent. In other words, a specific outcome of one event doesnโt change the number of possible outcomes of the other.

The fundamental counting principle says that if we have two events ๐ด and ๐ต and the number of possible outcomes for event ๐ด is ๐ and the number for event ๐ต is ๐, the total number of distinct possible outcomes of these two events together is ๐ times ๐. Letโs demonstrate how we can use the fundamental counting principle alongside the addition rule.

A cup contains 10 blue marbles, six green marbles, and seven red marbles. None of the marbles in the cup are identical. How many ways can four marbles be chosen from the cup so that exactly three of them are the same color?

Letโs begin by identifying the different events that will lead to an outcome of four marbles where exactly three of them are the same. We can select four marbles so exactly three of them are blue. We can select four where exactly three of them are green or four where exactly three of them are red. Now we notice that no outcomes are shared by the different events. So we say that these events are pairwise mutually exclusive.

In this case, we can use the addition rule. This says that the number of distinct outcomes from a collection of pairwise mutually exclusive events is the sum of the number of distinct outcomes from each event. So we need to calculate the number of outcomes in each event. Letโs start with the first, thatโs choosing three blue marbles. Essentially, in this case, weโre choosing three blue marbles and one that is not blue. Now, in fact, these two events are independent; a specific outcome of one doesnโt affect the number of possible outcomes of the other. And so we can use the fundamental counting principle. That tells us that the number of outcomes of the two events together is found by multiplying the number of outcomes from each event.

There are 10 blue marbles, and there are six plus seven marbles that are not blue. Thatโs 13. This means that there are 13 ways of choosing one marble that is not blue. Itโs a little bit more complicated when it comes to choosing three marbles from a total of 10. The order in which we choose these marbles doesnโt matter. And so we say that there are 10 ๐ถ or 10 choose three ways of choosing three marbles from a total of 10. The fundamental counting principle tells us then the number of ways to select four marbles so that exactly three of them are blue is 13 times 10 choose three.

We will now move on to the green marbles. There are six green marbles altogether and then 17 which are not green. So there are six choose three ways of choosing three green marbles from a total of six. And then by the fundamental counting principle, the number of ways of selecting four so that exactly three of them are green is 17 times six choose three.

Finally, we have the red marbles. Seven are red, and 16 are not red. That means there are seven choose three ways of choosing those three red marbles from a total of seven. There are 16 ways of choosing a marble thatโs not red. And so the fundamental counting principle says that the number of ways to select four marbles so that exactly three of them are red is 16 times seven choose three.

Finally, we apply the addition rule. So the number of different ways that four marbles can be chosen from the cup so that exactly three of them are the same color is the sum of these. Itโs 13 times 10 choose three plus 17 times six choose three plus 16 times seven choose three. Thatโs option (C).

Letโs look at one more example, but this time we wonโt use combinations.

Write the calculation we would use to work out the number of ways we can park two cars and then at least two trucks in five parking slots in a row. Is it five ๐ two times three ๐ three plus five ๐ two times three ๐ two, five choose two times three choose three plus five choose two times three choose two? Is it five ๐ two plus three ๐ three plus five ๐ two plus three ๐ two? Is it option (D) five choose two plus three ๐ three plus five ๐ two plus three ๐ two or option (E) five ๐ two times five ๐ three plus five ๐ two times five ๐ two?

Letโs identify the different events that will give us the described outcome. We could have two cars and three trucks parked, or we could have two cars and two trucks parked. And the events donโt share a common outcome, so theyโre mutually exclusive. The addition rule for two events says that if the events are mutually exclusive, then the number of outcomes of those two events is the sum of the number of distinct outcomes from each event.

So letโs find the number of outcomes. Letโs begin with the event where we park two cars and three trucks. Now we split this individually into two more events, that is, parking two cars and parking three trucks. Since a specific outcome of one event doesnโt affect the number of possible outcomes of the other, they are independent events. More specifically, if we park two cars in any parking slots, there are still three slots for the trucks to park.

And so we can use the fundamental counting principle. The number of outcomes of the two events together is the product of the number of outcomes from each event. Now since the order in which we park the cars and the order in which we park the trucks does matter, weโre talking about permutations. The number of ways of choosing ๐ items from a total of ๐ items when order does matter is ๐P๐.

So in this case, the number of ways to park the two cars in the five possible parking slots is five ๐ two. Once weโve parked the cars, there are only three possible parking slots left. So there are three different ways to organize the three trucks in those remaining spots. Thatโs three ๐ three. Since these events are independent, the fundamental counting principle tells us that the total number of outcomes is the product of these. So itโs five ๐ two times three ๐ three.

Letโs now repeat this process for two cars and two trucks. The first part of this is the same. Itโs five ๐ two. Weโre choosing two slots out of a total of five. But then the number of ways of organizing the two trucks into the remaining three spaces is three ๐ two. The total number of outcomes by the fundamental counting principle is the product of these. Itโs five ๐ two times three ๐ two.

Now since parking two cars and three trucks is mutually exclusive with parking two cars and two trucks, the total number of outcomes altogether, the number of ways in which we can park two cars and then at least two trucks in five parking slots, is found by adding these two expressions together. So itโs five ๐ two times three ๐ three plus five ๐ two times three ๐ two. And we can see thatโs option (A).

Letโs now recap the key points from this lesson. In this video, we learned that if ๐ด and ๐ต are mutually exclusive events where ๐ด has ๐ distinct outcomes and ๐ต has ๐ distinct outcomes, there are a total of ๐ plus ๐ distinct outcomes from either ๐ด or ๐ต. And this rule can be extended to apply to more than two pairwise mutually exclusive events. We also saw that it can be applied alongside the fundamental counting principle to solve more complicated problems.

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