Lesson Video: Combination Mathematics • Second Year of Secondary School

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 𝑛𝐶𝑘 represents the number of different ways to select 𝑘 objects out of a total of 𝑛 distinct objects. When dealing with combinations, the order does not matter. It is important to note that 𝑛 and 𝑘 must be nonnegative integers and 𝑛 must be greater than or equal to 𝑘. We recall that the permutation 𝑛𝑃𝑘 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 𝑛 is the total number of distinct objects and 𝑘 is the number of objects being ordered or selected. Let’s imagine a race where we have 𝑛 runners. And in the first scenario, the top 𝑘 runners receive medals with their ranks. If 𝑘 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 𝑘 objects from 𝑛 total objects. This is given by the permutation 𝑛𝑃𝑘. 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 𝑛𝐶𝑘. 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 𝐴 and 𝐵 such that the number of possible outcomes for event 𝐴 is 𝑥 and the number of possible outcomes for event 𝐵 is 𝑦, then the total number of distinct possible outcomes of these two events together is the product 𝑥 multiplied by 𝑦. Applying this to our example, we see that the number of ways to select 𝑘 runners from 𝑛 multiplied by the number of ways to order 𝑘 runners is equal to the number of ways to order 𝑘 runners from 𝑛. Using our notation for combinations and permutations, the first term on the left-hand side is 𝑛𝐶𝑘 and the term on the right-hand side is 𝑛𝑃𝑘.

As there are 𝑘 factorial ways to order 𝑘 runners, we have 𝑛𝐶𝑘 multiplied by 𝑘 factorial is equal to 𝑛𝑃𝑘, recalling that 𝑘 factorial is the product of all positive integers less than or equal to 𝑘. Dividing both sides of our equation by 𝑘 factorial, we have 𝑛𝐶𝑘 is equal to 𝑛𝑃𝑘 divided by 𝑘 factorial. Using the fact that 𝑛𝑃𝑘 is equal to 𝑛 factorial over 𝑛 minus 𝑘 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 𝑛𝐶𝑘 is equal to 𝑛 factorial divided by 𝑛 minus 𝑘 factorial multiplied by 𝑘 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 𝐶 five? Is it (A) 41 𝑃 five divided by five factorial, (B) 41 𝑃 five divided by five, (C) 41 𝑃 five multiplied by five factorial, or (D) 41 𝑃 five multiplied by five?

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

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

Evaluate 23 𝐶 19.

In order to answer this question, we begin by recalling the formula for combinations. 𝑛𝐶𝑘 is equal to 𝑛 factorial divided by 𝑛 minus 𝑘 factorial multiplied by 𝑘 factorial. In this question, 𝑛 is equal to 23 and 𝑘 is equal to 19. 23 𝐶 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 𝐶 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 𝑛𝐶𝑘 represents the number of different ways to select 𝑘 objects from 𝑛 distinct objects. This can be calculated by dividing 𝑛 factorial by 𝑛 minus 𝑘 factorial multiplied by 𝑘 factorial. In this question, we have a deck of 52 cards, so 𝑛 is equal to 52, and we wish to select three of them, so 𝑘 is equal to three. This means we need to calculate 52 𝐶 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 𝐶 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 𝑛 𝐶 three equals 120, find 𝑛.

We begin by recalling the formula for combinations. 𝑛𝐶𝑘 is equal to 𝑛 factorial divided by 𝑛 minus 𝑘 factorial multiplied by 𝑘 factorial. In this question, we are told that 𝑘 is equal to three, and we also know that 𝑛 𝐶 three is equal to 120. Substituting these values into our formula, we have 𝑛 factorial divided by 𝑛 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 𝑛 factorial over 𝑛 minus three factorial is equal to 720.

𝑛 factorial can be rewritten as shown. We can then divide the numerator and denominator of the left-hand side by 𝑛 minus three factorial. This leaves us with the equation 𝑛 multiplied by 𝑛 minus one multiplied by 𝑛 minus two is 720. The product of three consecutive integers is equal to 720. We could work out the value of 𝑛 using trial and improvement. However, we can also use the fact that for any 𝑛 greater than two, the product of these three integers is greater than 𝑛 minus two cubed and less than 𝑛 cubed. In this question, 720 is greater than 𝑛 minus two cubed and less than 𝑛 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 𝑛 minus two and less than 𝑛. Solving the two parts of this inequality, we see that 𝑛 must be less than 10.96 and it must be greater than 8.96. This means that we have two possible values for 𝑛: nine or 10. Substituting 𝑛 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 𝑛 equals 10, we have 10 multiplied by nine multiplied by eight. This is equal to 720. We can therefore conclude that if 𝑛 𝐶 three equals 120, then 𝑛 equals 10.

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

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