Video Transcript
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 π two times three π
three plus five π two times three π two, five choose two times three choose three
plus five choose two times three choose two? Is it five π two plus three π
three plus five π two plus three π two? Is it option (D) five choose two
plus three π three plus five π two plus three π two or option (E) five π two
times five π three plus five π two times five π two?
Letβs 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 donβt share a common
outcome, so theyβre 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 letβs find the number of
outcomes. Letβs 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 doesnβt 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, weβre talking
about permutations. The number of ways of choosing π
items from a total of π items when order does matter is πPπ.
So in this case, the number of ways
to park the two cars in the five possible parking slots is five π two. Once weβve 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. Thatβs three π three. Since these events are independent,
the fundamental counting principle tells us that the total number of outcomes is the
product of these. So itβs five π two times three π
three.
Letβs now repeat this process for
two cars and two trucks. The first part of this is the
same. Itβs five π two. Weβre 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 π two. The total number of outcomes by the
fundamental counting principle is the product of these. Itβs five π two times three π
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 itβs five π two times three π
three plus five π two times three π two. And we can see thatβs option
(A).