Lesson video

Hello there.

You made a great choice with today's lesson.

It's gonna be a good one.

My name is Dr.

Ronson and I'm gonna be supporting you through it.

Let's get started.

Welcome to today's lesson from the unit of conditional probability.

This lesson is called combinations.

And by the end of today's lesson, we will be able to derive the product rule for counting, including knowing when to divide by two.

Here are some previous keywords that will be useful during today's lesson, so you may want to pause the video if you need to remind yourself what any of them mean and then press play when you're ready to continue.

The lesson is broken into two learning cycles.

I'm going to start by looking at combinations when selecting one item per group.

Let's begin with this scenario here.

A restaurant has a two course menu, starter and a main.

Customers choose one starter and one main course for their meal.

How could we work out the number of possible combinations for the meal? Let's take some suggestions.

Aisha says, "we could write all the combinations as a list." Jacob says, "we could use a table." And Alex says, "we could use a tree diagram." These are all great suggestions.

Let's look at 'em one at a time.

Aisha lists all the possible combinations.

She says, "I'll start by writing all the combinations that include soup as a starter." There are three possibilities for this.

Soup with beef pie, soup with cheese and onion pie, and soup with ham and leek pie.

She then says, "I'll write all the combinations that include pate as a starter." And there are three possibilities for this too.

Pate with beef pie, pate with cheese, onion pie, and pate with ham leek pie.

So now we can see all the combinations.

We can see that there are six possible combinations of meals.

Let's take a look now at Jacob's method, he displays all the possible combinations as a two-way table.

He says, "I'll display the starters in rows and the main courses in columns" and then we can fill it in the cells of this table to see all the possible combinations of meals.

And once again, we can see that there are six combinations of meals.

And what Jacob has written inside his table is the same as what I shed written.

And then we have Alex's method.

He displays all the possible combinations as a tree diagram.

He says, "there are two possibilities for the starter, soup and pate." And for each of these starters there are three possibilities for the main course.

And then when we follow each of these branches, we can create each of the combinations.

Soup with beef pie, soup with cheese onion pie, soup with ham and leek pie, and then pate with each of those main courses as well.

So we can see here that there are six possible combinations of meals.

Let's compare these three methods now side by side.

All three of these methods show that there are six possible combinations of meals.

They also all show that for each of the two starters, there are three possibilities for the main course.

So what would change if the main course had four options rather than three? How would we adapt each of these representations and perhaps think about how many possible combinations the meals there would be.

Pause the video while you think about this and press play when you're ready to continue together.

Let's take a look.

There are now eight possible combinations of meals and we can see how each of these three representations has been adapted to account for the additional main course.

In the list, we can see that there are two additional combinations of meals written.

One for each starter paired up with the new main course.

In the table, we can see that there's an additional column for that new main course and that creates two new cells underneath it for each of the new combinations of meals.

And in the tree diagram, we can see that at the end of the branch for each starter, there's an additional branch now written for the main course as well.

So these representations all show that for each of the two starters, there are four possibilities for the main course.

What would change if the starter had three options? Perhaps pause the video while you think about how this might affect each of these representations and how many combinations there would be.

Pause the video while do that and press play when you're ready to continue together.

Let's take a look.

There would now be 12 possible combinations of meals.

And let's see how each of these representations has been adapted to account for the new starter.

In the list, we can see that there are four additional combinations written.

They are one for each of the main courses paired up with a new starter.

In the table, we can see that there's an additional row written.

And next to that it has four additional cells, one for each of the combinations that it creates.

And in the tree diagram in the first layer of branches, there is an additional branch for the asparagus starter.

And off that we've drawn four more branches for the main courses that could be paired up with that.

Each of these representations show that for each of the three starters, there are four possibilities that the main course.

Let's compare these three menus now side by side and consider the number of combinations that can be derived from each scenario.

With the menu on the left, there are two possible starters, three possible main, and that creates six possible combinations of meals.

For the menu in the middle, there were two possible starters, four possible main, and that creates eight possible combinations of meals.

And for the menu on the right, there are three possible starters, four possible mains, and that creates 12 combinations of meals.

Now we work these out by writing down all the combinations of meals and then counting them.

But let's see if we can work out how we might calculate the number of possible combinations based on the number of options for each course.

The number of combinations can be calculated by multiplying the number of possibilities for each course.

So for the menu on the far left, we can see that for each of the two starters, there are three main courses that could be paired up with them.

And that creates six combinations by doing two multiplied by three to get six.

For the menu in the middle, for each of the two starters, there are four main courses.

So that creates eight combinations by doing two multiplied by four.

And then for the menu on the right, for each of the three starters, there are four main courses that could be paired up with them.

And that creates 12 combinations by doing three multiplied by four.

This is sometimes referred to as a product rule for counting.

So let's check what we've learned.

Here we have two tables with children.

There are four children sat around one table and six children sat around another table.

The teacher chooses one child from each table.

In how many different ways could the teacher do this? Pause video while you write down your answer and press play when ready to see what it is.

The answer is 24, which we get by multiplying four by six.

The number of options from table one multiplied by the number of options at table two.

In other words, for every person that the teacher could pick from table one, there are six people that could be paired up with him from table two and that creates 24 combinations.

Let's now take a look at another restaurant menu, which has three courses this time.

Customers choose one starter, one main course, and one dessert for their meal.

How could we work out the number of possible combinations for their meal this time? Let's take some suggestions again.

Jacob says, "I can't use a two-way table for three courses." That's because it's a two-way table and not a three-way table.

You need to construct a three-way table in 3D, which is pretty hard to do.

Alex says, "I could use a tree diagram but it would be messy." Aisha says, "I could write a list but it would be long." So writing down all the combinations and then counting them might not be the most easy thing to do in this situation.

But could we do something different? They all say we could calculate the number of combinations by using the product rule for counting.

So we have three possible starters.

We have four possible main courses and we have two possible desserts.

If we multiply all those together, we do three times four times two, and that gives 24 combinations of meals altogether.

So let's check what we've learned with that.

Now we have three tables of children.

There are two sat around one table, three sat around another table, and five sat around another table.

A teacher chooses one child from each table.

In how many different ways could the teacher do this? Pause video while you write down an answer and press play when you're ready to see what the answer is.

The answer is 30, which we get from doing two multiplied by three multiplied by five.

In other words, we are multiplying the number of options from each table to get 30.

Okay, it's over to you now for task A, this task has two questions and here is question one.

Pause the video while do this and press play when you're ready for question two.

And here is question two, pause the video while you do this and press play when you're ready to go through some answers.

Let's see how we go on.

For question one, there are four possible outcomes when you draw a card from deck A.

There are five possible outcomes when you draw a card from deck B and there are 20 possible outcomes when you draw one card from each deck.

For each of the four possibilities from deck A, there are five possibilities from deck B and that creates 20 combinations of outcomes.

And then for question two, we have a restaurant menu.

And for part A, there are 20 different ways that a customer could choose one starter and one main course because there are four options for the starter and for each of those there are five options for the main course.

For part B, there are 10 different ways that a customer could choose a main course and a dessert because there are five options for main.

And for each of those there are two options that could be paired up with it for the dessert.

And there are 40 different ways that a customer could choose one item from each course.

And Part D, the restaurant runs out lasagna.

So in how many different ways could a customer choose one start and one main course and one dessertnow? There would be 32 different ways 'cause now we still have four possibilities for starters, but we only have four possibilities for mains and we still have two possibilities for desserts.

Four multiplied by four multiplied by two gives 32.

Well done so far.

Now let's move on to the next part of this lesson where we're going to look at other problems that involve combinations.

Let's begin with this scenario here where we have Sofia and Lucas who play against each other in a tennis match and they each choose one outta the five children we can see here to be on their team.

In how many different ways could the children be chosen? Well, this scenario could be considered as two separate trials.

Sofia's choice and Lucas's choice.

So based on that, Lucas says, "there are five possible outcomes for each choice." So if there are five possible outcomes for each choice, that means we can do five multiplied by five to say that there are 25 possible choices.

In other words, Sofia has five options and Lucas has five options.

So there are 25 options altogether.

But actually this isn't correct.

Why is this not quite correct? Perhaps pause a video and think about what is wrong with Lucas's reasoning here and press play when you're ready to continue together.

Let's take a look at this together now by going back to an old trusted method of writing out all the possible combinations.

All of the possible outcomes can be seen in this table on the screen here, the columns show us Lucas's choice and the rows show us Sophia's choice.

Can we now see the problem in this context? Lucas says, "in this context, the same person cannot be chosen twice.

For example, we can't both have Andeep on our team." So we can't count AA as one of the combinations.

Lucas says, "I had calculated 25 possible choices, but this would only be true if the same choice could be repeated.

There are actually 20 possible choices." Because outta those 25 choices we had, five of them were ones where we had the same person chosen twice.

Therefore there are only 20 combinations altogether once we discount those five.

So let's work through the scenario again now, but in a slightly different way.

In particular, let's think about how one person will make the first choice and then this next person will make a choice on whoever is left over.

So there are five options for whoever makes the first choice, which means the possible outcomes of the first choice is five.

But once that first person is chosen, there are only four options remaining for whoever makes the second choice.

So the number of possible outcomes for the second choice is only four.

That means for each of the five possible outcomes of the first choice, there are four possible outcomes for the second choice.

For example, if Andeep is chosen first, there are four options remaining for whoever chooses second, Laura, Izzy, Sam, and Jun.

But if Laura is chosen first, there are four options remaining for the person who chooses second, but they are Andy, Izzy, Sam and Jun.

And then same if Izzy is chosen first, there are four options remaining.

And if Sam has chosen first, there are four options remaining and if Jun is chosen first, there are four options remaining.

So that means there are 20 possible choices altogether because for each of the five first choices, there are four second choices remaining.

So let's check what we've learned.

Here, we have Alex who has a pack of six cards.

They are labelled A to F.

He deals one card to himself and deals one card to Jacob.

In how many different ways can this happen? Pause video while write down your answer and press play when you're ready to see what it is.

The answer is 30, which we get by multiplying six by five.

That's because there are six possibilities for what card Alex could get.

But once he's got a card, there are only five possibilities left for what card Jacob could get.

Let's change the scenario ever so slightly here.

Alex picks a card at random, he then replaces the card, in other words, he puts it back in the pack and then picks a card at random again.

In how many different ways can this happen? Pause video while you write down your answer and press play when you're ready to see what it is.

This time the answer is 36.

It's done by doing six multiplied by six.

The reason why we are multiplying six by six is because there are six options for Alex's first card, but then because he puts the card back in again, there are still six options for the second card he draws.

So altogether there are six different possibilities.

Let's now look at a different variation of this tennis scenario.

Sofia and Lucas this time play on the same team in a tennis match and Sofia chooses two out of the five children below to be on the opposing team.

In how many different ways could the children be chosen? Well, once again, this scenario could be considered as two separate trials.

It would be in this case Sofia's first choice and Sofia's second choice.

And the context suggests that the two trials could not have the same outcome.

In other words, the same person can not be chosen twice.

Sofia says, "for each of the five possible outcomes in the first choice, there are four possible outcomes in the second choice." Which means the number of possible pairs that can be chosen to play against Sofia and Lucas will be 20.

That's from doing five multiplied by four.

But once again, this isn't quite correct.

Can you think about why this is not quite correct? In particular, think about the context here.

Pause the video while you think about this and press play when you're ready to continue together.

Well, let's think about this by considering all of the outcomes written out again.

Here we have a two-way table that shows all the possible outcomes.

The row show us Sofia's first choice and the column show us Sofia's second choice.

And we can see that all the cells that would include the same person being chosen twice have been blacked out.

So they are not counted here.

Sofia says, "in this context, the order of my choices doesn't matter.

So the outcomes AI and IA both result in the same pair of children being chosen." They both result in Andeep and Izzy being a team, that means they are in fact the same outcome for this particular context.

Sofia said, "I calculated 20 possible pairs, but this would only be true if the order mattered.

However, the table shows that I've counted each pairing twice." So here we can see that the pairing of Andeep and Izzy has been counted twice.

So is Andeep and Jacob and Andeep and Lucas and Andeep and Sam.

So Sofia says, "the fact that the order doesn't matter means that there are only actually half as many different possible pairs than I had previously counted." And so it means there are 10 possible pairs that Lucas and Sophia could play against.

When the order of outcomes does not matter, we can calculate the number of possible combinations by multiplying the number of possible outcomes per trial and then dividing by two.

For example, in this situation there were five possible outcomes to the first choice.

There were then only four possible outcomes for the second choice.

But because the order didn't matter, it meant the number of possible pairs that could be chosen was five multiplied by four then divided by two which gives 10.

So let's check what we've learned.

Here we have Alex who has a pack of six cards.

He chooses two cards at random from the pack.

In how many different ways can he do this? Pause video while you write down your answer and press play when you're ready to see what the answer is.

Well, this question doesn't really make a distinction between which one is the first card and which one's the second card.

It just talks about which cards could he end up with at the end of the process.

He does draw two cards one at a time.

So that means there are six options for the first and then five options for the second.

But because the order doesn't matter, we then divide by two to get 15.

Okay, it's over to you now for task B.

This task has six questions and here are questions one to four.

Pause the video while you do these and press play for more questions.

Here is question five, pause video while you do this and press play for question six.

And here is question six.

Pause video while you do this and press play to go through some answers.

Okay, let's go through some answers.

In question one, there are 52 cards in a pack.

In part A Lucas draws one card from the pack, replaces it and draws another card from the pack.

Here we are making a distinction between the first card he draws and the second card.

And also we notice that there are 52 choices each time for Lucas, because he puts the card back in the pack.

Which means we do 52 multiplied by 52 to get 2,704.

In part B, Aisha draws two cards in the pack at once.

So this time doesn't really matter what order she draws those cards in.

For the sake of the calculations to begin with, you should maybe think about the fact she does choose one card and then the other.

So it's 52 multiplied by 51, but at the end of the process, it doesn't really matter what order those are in.

So we divide by two to get 1,326.

Then question two, Izzy and Jun each flip a fair coin and you have to work out the number of different combinations of outcomes.

Well, there are two possibilities for Izzy, heads and tails.

There are two possibilities for Jun, heads and tails and they could all be pair of each other.

So the answer is two multiplied by two, which is four.

I could think of it as two to the power of two.

In question three, it's the same scenario, but there are three children flipping a coin.

Each child has two possible outcomes, which means the number of outcomes altogether are two multiplied by two multiplied by two, which is the same as doing two to power of three.

Two to the power of the number of children and that'll be eight.

And question four, it's a similar situation again, but this time you've got four children who are each rolling a regular six-sided dice.

So number of different combinations would be six for Alex, six for Jacob, six for Laura, and six for Sofia all multiplied together.

Or you could even think of it as six, which is the number of outcomes each time, to the power of four, which is the number of children, and that would be 1,296.

In question five you had two sets of cards, A and B.

In part A, Andeep draws one card from set A and one card from set B.

There are 50 different ways of doing that.

In part B, he draws two different cards from set A.

Well, this time there would be five multiplied by four because there are five options for his first choice, four for his second choice and divided by two because it doesn't really make a distinction between which one is the first and second and that'd give 10 possible outcomes.

In C, Laura draws two different cards from set B.

Do the same process again, but it's 10 multiplied by nine divided by two to get 45.

In part D, Alex draws two different cards from set A and two different cards from set B, well, to get the total number of combinations, you can think about your previous answers.

There are 10 different combinations for the two cards that can be drawn from set A.

And for each of those there are 45 different combinations for the two cards that can be chosen from set B, which means we can do 10 multiplied by 45 to get 450.

Then question six was all about choosing a safe password.

Jun's password contains three characters and each one is a lowercase letter.

So that means there are 26 possibilities for each character.

And there are three characters.

So you knew 26, so the power of three.

Or you can multiply 26 by 26 by 26 if you want to.

That will give 17,576 possible combinations for his password.

Izzy's password contains three characters.

Each character is a lowercase letter or an uppercase letter.

That means there are 52 possible outcomes for each character.

That is two possible outcomes each letter, whether it's lowercase or upper case, and as there are three characters, you can do 52 to the power of three to get 140,608.

Sam's password contains three characters.

Each character is either a lowercase letter, an uppercase letter, or a numeral.

Well, the number of possibilities for each character would be the 26 lower case letters plus the 26 upper case letters, plus the 10 possible numerals, which means you can do 62 to the power three to get 238,328 possible passwords.

And then Sofia's password contains three characters.

Each character is either a lowercase letter, an uppercase letter, a numeral, or one of the symbols below.

Well, this time there are 77 possibilities for each character.

There are the same 62 as what Sam had, plus the additional 15 characters we can see from these symbols on the screen.

We can then do 77 to the power of three to get 456,533 possible passwords.

And then Aisha's password contains eight characters.

Each character is either a lowercase letter, an uppercase letter, a numeral or one of the symbols above.

In this case, once again, there are 77 possibilities for each character, and this time there are eight characters.

So we can do 77 to the power of eight, which gives you, well that very large number you can see on the screen there, it is over a quadrillion different possibilities for Aisha's password.

It's approximately 1.

24 times 10 to the power of 15 in terms of the number of possibilities for Aisha's password.

So you could say Aisha's password is the safest one on there.

It's the hardest one to guess by going through all the possibilities.

Fantastic work today.

Now let's summarise what we've learned.

By systematically listing the outcomes for two or more events, patterns begin to emerge and we spotted some of those patterns during today's lesson.

The pattern can be generalised so that counting possible outcome combinations is quicker.

For example, the product rule for counting.

The context behind a problem though can affect what calculations you perform.

For example, whether the outcome can be the same or not each time, or whether the order of the outcomes matter as well.

So it's always worth considering the context behind the calculations you are doing.

Well done today.

Have a great day.