### Video Transcript

In this video, we will learn how to
use combination properties to solve problems and use combinations to count possible
outcomes. The combination ππΆπ represents
the number of different ways to select π objects out of a total of π distinct
objects. When dealing with combinations, the
order does not matter. It is important to note that π and
π must be nonnegative integers and π must be greater than or equal to π. We recall that the permutation
πππ is very similar. However, in this case, order does
matter. We will begin this video by
comparing permutations and combinations in context.

As already mentioned, the key
difference between a permutation and a combination is whether the order matters. The notation used for both of these
is similar, where π is the total number of distinct objects and π is the number of
objects being ordered or selected. Letβs imagine a race where we have
π runners. And in the first scenario, the top
π runners receive medals with their ranks. If π is equal to three, two
possible ways to assign medals are as shown. Even though the same three runners
took the top three spots in both cases, they resulted in the medalists winning
different medals.

This is because order matters. The number of different ways to
assign medals for this race is the number of different ways to order π objects from
π total objects. This is given by the permutation
πππ. If, on the other hand, the top
three finishers were given identical medals, as opposed to those with their
positions on, then there is only one way of selecting the same three finishers. This is because the order does not
matter. This value is given by the
combination ππΆπ. We will now consider how we can
calculate this using the fundamental counting principle.

The fundamental counting principle
states that if we have two independent events π΄ and π΅ such that the number of
possible outcomes for event π΄ is π₯ and the number of possible outcomes for event
π΅ is π¦, then the total number of distinct possible outcomes of these two events
together is the product π₯ multiplied by π¦. Applying this to our example, we
see that the number of ways to select π runners from π multiplied by the number of
ways to order π runners is equal to the number of ways to order π runners from
π. Using our notation for combinations
and permutations, the first term on the left-hand side is ππΆπ and the term on the
right-hand side is πππ.

As there are π factorial ways to
order π runners, we have ππΆπ multiplied by π factorial is equal to πππ,
recalling that π factorial is the product of all positive integers less than or
equal to π. Dividing both sides of our equation
by π factorial, we have ππΆπ is equal to πππ divided by π factorial. Using the fact that πππ is equal
to π factorial over π minus π factorial, we will rewrite the right-hand side so
we have a formula for combinations. The formal definition of a
combination is as shown, where ππΆπ is equal to π factorial divided by π minus π
factorial multiplied by π factorial.

It is worth noting that there are
equivalent notations, including the two shown. We will now look at a couple of
examples to reinforce this.

Which of the following is equal to
41 πΆ five? Is it (A) 41 π five divided by five
factorial, (B) 41 π five divided by five, (C) 41 π five multiplied by five
factorial, or (D) 41 π five multiplied by five?

We begin by recalling our formulae
for permutation and combination. The permutation πππ is equal to
π factorial divided by π minus π factorial. And the combination ππΆπ is equal
to π factorial divided by π minus π factorial multiplied by π factorial. Combining these two formulae, we
see that ππΆπ is equal to πππ divided by π factorial. In this question, π is equal to 41
and π is equal to five. This means that 41 πΆ five is equal
to 41 π five divided by five factorial. The correct answer is therefore
option (A).

In our next example, we need to
evaluate a combination.

Evaluate 23 πΆ 19.

In order to answer this question,
we begin by recalling the formula for combinations. ππΆπ is equal to π factorial
divided by π minus π factorial multiplied by π factorial. In this question, π is equal to 23
and π is equal to 19. 23 πΆ 19 is therefore equal to 23
factorial divided by 23 minus 19 factorial multiplied by 19 factorial. The denominator simplifies to four
factorial multiplied by 19 factorial. We can rewrite the numerator as 23
multiplied by 22 multiplied by 21 multiplied by 20 multiplied by 19 factorial. And by dividing the numerator and
denominator by 19 factorial, we have 23 multiplied by 22 multiplied by 21 multiplied
by 20 divided by four factorial.

Writing four factorial as four
multiplied by three multiplied by two multiplied by one, we can reduce the factors
as shown. This leaves us with 23 multiplied
by 11 multiplied by seven multiplied by five, which is 8,855. 23 πΆ 19 is equal to 8,855.

In our next example, we will
examine a counting problem involving combinations.

How many three-card hands can be
chosen from a deck of 52 cards?

We begin by recalling that the
combination ππΆπ represents the number of different ways to select π objects from
π distinct objects. This can be calculated by dividing
π factorial by π minus π factorial multiplied by π factorial. In this question, we have a deck of
52 cards, so π is equal to 52, and we wish to select three of them, so π is equal
to three. This means we need to calculate 52
πΆ three, which is equal to 52 factorial divided by 49 factorial multiplied by three
factorial.

Using our knowledge of factorials,
we know that 52 factorial can be rewritten as 52 multiplied by 51 multiplied by 50
multiplied by 49 factorial. This allows us to cancel 49
factorial on the numerator and denominator. Three factorial is equal to three
multiplied by two multiplied by one. We can then cancel factors of three
and two as shown. 26 multiplied by 17 multiplied by
50 is equal to 22,100. As 52 πΆ three is equal to 22,100,
this is the number of three-card hands that can be chosen from a deck of 52
cards.

In our final example, we will
consider how to identify unknown parameters in combinations.

If π πΆ three equals 120, find
π.

We begin by recalling the formula
for combinations. ππΆπ is equal to π factorial
divided by π minus π factorial multiplied by π factorial. In this question, we are told that
π is equal to three, and we also know that π πΆ three is equal to 120. Substituting these values into our
formula, we have π factorial divided by π minus three factorial multiplied by
three factorial is equal to 120. Using the fact that three factorial
is equal to six, then multiplying both sides of our equation by three factorial
gives us π factorial over π minus three factorial is equal to 720.

π factorial can be rewritten as
shown. We can then divide the numerator
and denominator of the left-hand side by π minus three factorial. This leaves us with the equation π
multiplied by π minus one multiplied by π minus two is 720. The product of three consecutive
integers is equal to 720. We could work out the value of π
using trial and improvement. However, we can also use the fact
that for any π greater than two, the product of these three integers is greater
than π minus two cubed and less than π cubed. In this question, 720 is greater
than π minus two cubed and less than π cubed. We can then cube root each part of
the inequality.

To two decimal places, the cube
root of 720 is 8.96. Therefore, this is greater than π
minus two and less than π. Solving the two parts of this
inequality, we see that π must be less than 10.96 and it must be greater than
8.96. This means that we have two
possible values for π: nine or 10. Substituting π equals nine into
the left-hand side of our equation, we have nine multiplied by eight multiplied by
seven. This is equal to 504, which is not
correct as the right-hand side is 720. When π equals 10, we have 10
multiplied by nine multiplied by eight. This is equal to 720. We can therefore conclude that if
π πΆ three equals 120, then π equals 10.

We will now finish this video by
recapping the key points. The combination ππΆπ represents
the number of different ways to choose π objects from π total distinct objects,
where the order of the π objects does not matter. The three different notations
written are all equivalent. The permutation πππ represents
the number of different ways to order π objects from π total objects. This time, the order of the π
objects does matter. We saw that ππΆπ is equal to π
factorial divided by π minus π factorial multiplied by π factorial. And we recall that πππ is equal
to π factorial divided by π minus π factorial.

This led us to the fact that the
number of combinations is equal to the number of permutations divided by π
factorial. Whilst we did not cover it in this
video, the combination ππΆπ satisfies the identity ππΆπ is equal to ππΆπ minus
π. For example, 12 πΆ three is equal to
12 πΆ nine as 12 minus three is equal to nine.