### Video Transcript

In this video, weโll 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. Thatโs 10 plus seven, which is
equal to 17 different options for lunch.

Now, of course, this is only true
because thereโs no common lunch item thatโs sold in both restaurants. If the Soup Kitchen, for instance,
decided that they were going to suddenly sell pepperoni pizza, weโd need to rethink
our calculations somewhat.

Letโs put this in general
terms. Remember, a pair of events is said
to be mutually exclusive if thereโs 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 ๐ด and ๐ต
are mutually exclusive and there are ๐ distinct outcomes of event ๐ด and ๐
distinct outcomes of event ๐ต, the total number of outcomes is ๐ plus ๐.

Now weโre going to want to apply
this alongside the rules we have for combinations and permutations. A combination is found when weโre
selecting objects from a large group and order doesnโt matter. So say we had three objects ๐ด, ๐ต,
and ๐ถ. Choosing ๐ด then ๐ต would be the
same as choosing ๐ต then ๐ด. And to find the number of ways of
choosing ๐ items from ๐ when order doesnโt matter, we find ๐C๐. When order does matter, itโs a
permutation. And the number of ways of choosing
๐ items from ๐ in this case is ๐P๐.

So with all this in mind, letโs
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 ๐ถ three times six ๐ถ
two? Is it (B) 10 ๐ถ three plus six ๐ถ
two? Is it (C) 10 ๐ three times six ๐
two, (D) 10 ๐ three plus six ๐ two, or (E) 10 ๐ถ three minus six ๐ถ two?

Weโre forming a group that consists
of either three boys or two girls. So in fact, there are two
events. The first event, letโs call that
event ๐ด, is the event that we choose three boys from a total of 10. The second event, letโs call that
๐ต, 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 weโre going to
be able to apply the addition rule to answer this problem.

This says that if two events ๐ด and
๐ต are mutually exclusive, where ๐ด has ๐ distinct outcomes and ๐ต 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.

Weโre choosing three boys from a
total of 10. Now thereโs no indication here that
order matters. In fact, letโs 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
doesnโt actually matter. We still end up with the same final
three boys. And so weโre dealing with
combinations. The number of combinations there
are, which is when order doesnโt matter, of choosing ๐ items from ๐ is ๐C๐ or ๐
choose ๐.

And the notation that weโve used
here will not necessarily be the notation youโre 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, weโll
calculate the number of ways of choosing three boys from a total of 10. Itโs 10 choose three. Similarly, the order in which the
girls are chosen doesnโt matter. And weโre choosing two from a total
of six. So itโs 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. Letโs now see how we can generalize
this to apply it to more than two.

Returning to our problem where
weโre 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. Letโs 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 ๐ถ four times six ๐ถ four
times seven ๐ถ 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 ๐ถ four plus six ๐ถ four plus seven ๐ถ four, or (E)
10 ๐ถ four times six ๐ถ four plus seven ๐ถ four?

Weโre 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, weโre
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 weโll add these values
together.

Now since the order in which these
balls are chosen doesnโt 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 doesnโt 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 ๐ถ four plus six ๐ถ four plus seven ๐ถ four. And thatโs option (D).

Now a powerful property of the
addition rule is that it can be used alongside the fundamental counting
principle. Thatโs sometimes also called the
multiplication rule or the productโs rule for counting. Whilst the addition rule, weโve
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 doesnโt change the number of possible outcomes of the other.

The fundamental counting principle
says that if we have two events ๐ด and ๐ต and the number of possible outcomes for
event ๐ด is ๐ and the number for event ๐ต is ๐, the total number of distinct
possible outcomes of these two events together is ๐ times ๐. Letโs 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?

Letโs 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. Letโs start with the first, thatโs
choosing three blue marbles. Essentially, in this case, weโre
choosing three blue marbles and one that is not blue. Now, in fact, these two events are
independent; a specific outcome of one doesnโt 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. Thatโs 13. This means that there are 13 ways
of choosing one marble that is not blue. Itโs 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 doesnโt matter. And so we say that there are 10 ๐ถ
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 thatโs 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. Itโs 13 times 10 choose three plus
17 times six choose three plus 16 times seven choose three. Thatโs option (C).

Letโs look at one more example, but
this time we wonโt 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?

Letโs 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 donโt share a common
outcome, so theyโre 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 letโs find the number of
outcomes. Letโs 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 doesnโt 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, weโre talking
about permutations. The number of ways of choosing ๐
items from a total of ๐ items when order does matter is ๐P๐.

So in this case, the number of ways
to park the two cars in the five possible parking slots is five ๐ two. Once weโve 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. Thatโs 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 itโs five ๐ two times three ๐
three.

Letโs now repeat this process for
two cars and two trucks. The first part of this is the
same. Itโs five ๐ two. Weโre 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. Itโs 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 itโs five ๐ two times three ๐
three plus five ๐ two times three ๐ two. And we can see thatโs option
(A).

Letโs now recap the key points from
this lesson. In this video, we learned that if
๐ด and ๐ต are mutually exclusive events where ๐ด has ๐ distinct outcomes and ๐ต has
๐ distinct outcomes, there are a total of ๐ plus ๐ distinct outcomes from either
๐ด or ๐ต. 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.