Lesson Video: Counting Outcomes with Restrictions Mathematics

In this video, we will learn how to count the number of possible outcomes when we have restrictions.

14:42

Video Transcript

In this video, we’re going to learn how to count a number of possible outcomes when given some restrictions. And to do so, we begin by recalling the fundamental counting principle, sometimes called the product rule for counting. It says if events 𝐴 and 𝐵 which are independent events have 𝑚 and 𝑛 possible outcomes, respectively, then the total number of possible outcomes for the events together is the product of these. It’s 𝑚 times 𝑛.

For instance, how many ways can a three-digit pin number be created by using the numbers zero to nine?

There are 10 possible numbers that we can use for each digit. That means there are 10 ways to choose the first digit, 10 ways to choose the second digit, and 10 ways to choose the third digit. The fundamental counting principle tells us that the total number of possible outcomes then is the product of these. It’s 10 times 10 times 10 or of course 10 cubed, which is equal to 1000. Now, in fact we can generalize this. And we can say that when counting with replacement, in other words, repetition is allowed, the total number of outcomes from 𝑛 repeated events of choosing from 𝑚 items is given as 𝑚 to the power of 𝑛.

Notice, though, that this is specifically for independent events, those where the outcome of the first event doesn’t affect the outcome of the second. In other words, in this case, when we chose that first digit, it didn’t affect the number that we could choose for the second digit. If this is not the case, for instance, where there are restrictions on using that number twice, then we can still use the fundamental counting principle. But we do need to be a little bit careful. Let’s see what that might look like.

How many four-digit numbers, with no repeated digits, can be formed using the elements of the set containing zero, one, three, and four?

So we’re creating numbers with four digits in them, and we’re not allowed to repeat those digits. We know that we can count the total number of possible outcomes by using the fundamental counting principle or the product rule for counting. This rule says that when combining more than one event, the total number of outcomes is found by multiplying the number of outcomes for each event together. And so we need to identify what each event actually is. The first event is picking the first digit, the second event is picking the second digit, and so on. There are four elements in our set, but this doesn’t mean there are four ways of choosing the first digit. In fact, for our number to be a four-digit number, its first digit cannot be zero. It can only be one, three, or four. And so there are actually just three ways of choosing the first digit.

Then we consider the second digit. We’ve already chosen a number from the list one, three, and four. And we know we cannot repeat one of these digits. And so if we take one of the numbers out of our set, we now have just three left. There are three ways of choosing the second digit. Now we move on to the third digit and we say that we’ve already chosen two possible numbers from the set, and that leaves us with two left. And similarly, when we get to the fourth digit, we’ve already taken three numbers, and so there’s only one left to choose from. The counting principle says then that the total number of outcomes, which is here the total number of four-digit numbers, is the product of these values. It’s three times three times two times one, which is equal to 18. We can make 18 four-digit numbers given that no digits can be repeated using the elements of our set.

Let’s consider a similar example to see if we can generalize this somehow.

David’s password must be five characters long. He can use the digits zero to nine and cannot use the same digit more than once. How many different passwords could David create?

We’re looking to find the total number of five-digit passwords, bearing in mind that we cannot use the same digit more than once. And so we recall the fundamental counting principle or the product rule for counting. This tells us that we can find the total number of outcomes for two or more events by multiplying the number of outcomes for each event together. Now, the events here are choosing the first character, choosing the second character all the way through to choosing the fifth character. And since we’re working with the digits zero to nine inclusive, there are a total of 10 digits to choose from. And there are therefore 10 possible different outcomes for the first event for choosing that first digit.

Now it’s really important that we realize we can’t use the same digit more than once when considering the total number of outcomes for our second event, that is, choosing the second digit. We’ve already chosen one from the digits zero to nine, and so that leaves us with just nine more to choose from. Similarly, when we get to the third digit, we know we’ve already taken two possible digits from our list. And so there are eight more to choose from.

In a similar way, there are seven ways of choosing the fourth digit and just six ways of choosing the fifth. The counting principle or the product rule for counting tells us that we now need to multiply these numbers together. 10 times nine times eight times seven times six is 30,240. And so we see there are 30,240 different passwords that David could create given that he can only use the digits zero to nine and cannot use them more than once.

Now we can actually generalize this result. We call this counting without replacement. And that is because we take a digit away, we don’t replace it and use it once again. And we say that when counting without replacement, the total number of ways of choosing 𝑛 items from a collection of 𝑚 is 𝑚 times 𝑚 minus one times 𝑚 minus two all the way down to 𝑚 minus 𝑛 minus one.

Let’s now consider another way of imposing restrictions.

After a recent reorganization, James is taking over responsibility for the manufacturing of odd numbers on the house sign number production line. As part of his scientific investigation into production levels, he wants to know how many three-digit numbers only contain odd digits. Calculate the answer for him.

We are interested in finding the total number of three-digit number house signs. However, there’s a pretty hefty restriction on this. These numbers can only contain odd digits; that is, they must be made up of the numbers one, three, five, seven, or nine. And so let’s consider each of the digits in turn.

The first digit can be any one of these numbers. It can be one, three, five, seven, or nine. So there are five possible ways of choosing the first digit. There are no restrictions on using the same digit more than once. For example, we could choose the number one, one, one; that would be fine. And so there are still five ways of choosing the second digit. The second digit can be any one of these odd numbers. Then, for the third digit, we have the exact same situation. We can choose the numbers one, three, five, seven, or nine. And so there are five ways of choosing the third digit.

The fundamental counting principle or the product rule for counting says the total number of ways of choosing these then is the product of these. It’s five times five times five, which is 125. There are 125 three-digit numbers then that only contain odd digits.

Let’s consider a slightly different context.

A building has five doors which are numbered as one, two, three, four, five. Determine the number of ways a person can enter and then leave the building if they cannot use the same door twice.

Let’s try to visualize this. Our building has five doors, and let’s label them one, two, three, four, five as it tells us to. Let’s imagine we have someone looking to enter the building. They have five possible ways to do so. But let’s imagine for sake of argument that they’re going to choose door number two. Once inside the building, we’re told they cannot use the same door twice, and so we cut off door two as an exit. Looking around, we now see that there are one, two, three, four possible ways for that person to exit the building. They may, for example, choose door four. There are therefore five possible ways to enter the building. But once we are into the building, there are only four possible ways to get out.

The product rule for counting or the counting principle tells us that the total number of ways a person can enter and then leave the building given these restrictions is the product of these. It’s five times four, which is equal to 20. There are 20 possible ways then that the person can enter and then leave the building given that they can’t use the same door twice.

In our very final example, we’re going to look at working out the number of possibilities for a seating arrangement.

Mia and Daniel are planning their wedding. They’re working on the seating plan for the top table at the reception. Their top table is a straight line with eight seats down one side. It needs to sit the bride and groom, the bride’s parents, the groom’s parents, the best man, and the maid of honor. Given that all couples need to sit next to each other and that the best man and maid of honor are not a couple, how many different ways are there for seating everyone on the top table?

We have a few restrictions on how we seat each couple and the maid of honor and the best man. Let’s begin by considering the couples who are the bride and groom, the bride’s parents, and the groom’s parents as three units essentially. And we’re going to begin by working out the total number of ways of just seating these three couples. There are three ways of choosing the first couple to seat. There are two ways of choosing the second couple to seat and one way of choosing the third. And, of course, the product rule for counting or the counting principle says the total number of options is the product of these. It’s three times two times one, which is six.

So we have six ways of seating the couples. But of course, each couple could sit in a different order. We could have the bride and groom or the groom then the bride. And if we think about it, there are two ways to seat the bride and groom, two ways to seat the bride’s parents, and two ways to seat the groom’s parents. Two times two times two is equal to eight, meaning that there are eight ways that each couple could sit next to each other. Bear in mind that this is for each of the six original ways of seating the couples. This means that the total number of possibilities when it comes to seating these is the product of these two sets of outcomes. It’s six times eight, which is 48. So we have 48 ways in total of seating those couples.

And so now we move on to seating the best man and the maid of honor. We consider these individually because we’re told they’re not a couple, and therefore they don’t necessarily need to sit next to one another. And so if we think about the top table with our three couples already seated, he could sit at either end. But also he could sit at any point between the couples. And so there must be four options of chairs for him. Then, once the best man is seated, the maid of honor could sit at either end. But she could also sit between any of the couples and/or the best man, depending on where he’s located. And this must mean that there are five different ways of seating the maid of honor.

Now that we’ve considered all the possible events, that is, seating the couples, seating the best man, and seating the maid of honor, we know that the fundamental counting principle tells us to find the product of these. That’s 48 times four times five, which is equal to 960. There are a total number of 960 different ways for seating everyone on the top table.

Now, in fact, this isn’t the only method of answering this problem. We can alternatively just consider that there are five different groups; there are three couples and two individuals. And so we would say that there are five ways of choosing the first group to seat, four ways of choosing the second group, three ways of choosing the third, and so on, giving us a total of 120 different ways to arrange these five places.

Then we go back to considering how the couples are arranged. We know that each couple could sit in a slightly different order. And so there are two times two times two, which is eight arrangements for our couples. Once again, the fundamental counting principle tells us that the total number of different ways for seating everyone is the product of these. It’s 120 times eight, which is once again 960.

We’re now going to consider the key points from this lesson. In this video, we saw that we can extend the use of the fundamental counting principle to scenarios where there are restrictions on the possible outcomes. We saw that when counting with replacement, the total number of outcomes from 𝑛 repeated events of choosing from 𝑚 items is 𝑚 to the power of 𝑛. When counting without replacement though, we have 𝑚 times 𝑚 minus one times 𝑚 minus two all the way down to 𝑚 minus 𝑛 minus one.

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