Counting: Videos & Practice Problems

Understanding permutations and combinations is essential in probability and statistics. Permutations account for the arrangement of distinct objects where order matters, calculated using the formula $n^{!}/{\mathrm{(n\; -\; r)}}^{!}$. In contrast, combinations group objects where order does not matter, represented by $n^{!}/r^{!}{\mathrm{(n\; -\; r)}}^{!}$. Mastering these concepts enhances problem-solving skills in various applications.

Introduction to Permutations

Hey, everyone. We just learned how to find the number of different ways we could wear different outfits on any given day. But what if instead of a single day, we wanted to look at the week ahead and determine the different ways we could wear five different shirts over five different days? Well, the number of ways that we could choose to do this is actually something called permutations, which are just a way to arrange a number of things in a specific order with each thing occurring only once. Now that word permutations might make this sound like it's going to be complicated and a bit tricky, but you don't have to worry about that because here I'm going to walk you through exactly what permutations are and how we can calculate them using the fundamental counting principle and factorials, two things that we already know how to do.

So let's go ahead and get started. When looking at our five different shirts that we want to figure out how to wear over five different days, we're going to use the fundamental counting principle and first look at the number of options that we have on that first day. So on Monday, I have all five different shirts to choose from, so I have five total options. But then on Monday, if I pick to wear this pink shirt, then on Tuesday, I'm left with only four options. So if on Tuesday, I choose to wear this teal shirt over here, on Wednesday, I now only have three options left, then two on Thursday and one on Friday.

Now this five times four times three times two times one, you might recognize as being a five factorial, and you'd be right. The number of different ways we could wear five different shirts over five days is equal to five factorial. But what if instead of five shirts, I actually had eight different shirts to choose from? How would I choose that? Well, on that first day, I now have eight total options because now I have eight shirts to choose from.

Then on that second day, I only have seven and then six and then five and then four. But this is not equal to eight factorial. So if we can't always find permutations by just taking the factorial, how exactly are we going to calculate it? Well, permutations have a specific formula, this formula right here, that is based on the number of total options that we're given and how many things we're picking out of that total. So if I take this formula and apply it to my problem with eight shirts over five days, I'm going to take the total number of options I'm faced with, in this case eight, and the factorial of that and divide it by that total minus the number of things that I'm choosing.

In this case, since I'm only choosing shirts for five days, my number here is five, then the factorial of that. Now this simplifies to 8!−3!, which is actually exactly what this here is. So a couple of things that I want to point out about our equation here is that n is always going to be the largest number. So n factorial on the top, that n is always going to be your total. It's going to be the largest number that you're given in a problem.

Then on the bottom, we're going to take that total and subtract our smaller number. Now in terms of notation, you may see this written as pnr here or pn,r in parentheses. Now this can be read as the number of permutations of n things taken r at a time. Now that might sound a little bit overwhelming just because there are so many words going on, but let's go ahead and work through some examples together to make this a bit more clear. So in our first problem here, we have a teacher that is choosing a line leader and a door holder from her class of 25 students.

So we want to know how many ways there are for the teacher to choose these positions. So using our permutations formula here, the first thing we want to do is identify our values for n and r. Now remember, n is always going to be your largest number here. So in this case, we have a total of 25 students, so this 25 represents our value of n. Now for r, we're looking at how many things we're choosing out of that total.

So we're choosing a line leader and a door holder, so that tells us that our value for r is equal to two because we're choosing two positions out of our total of 25 different students. So in that permutations notation, I would write that as pn=r25,2. And from here, we're just going to plug these values into our formula. So remember, our largest number, our n, is always going to come on the top there so that I have 25! on the top. Then on the bottom, I take that 25 and I subtract that smaller number, in this case two, and I take the factorial of that.

Now simplifying this, this gives me 25!−23!. Now you may be worried that we're now going to have to write all of these factorials out, but you don't have to worry about that because there's actually a much easier way to solve this by simply rewriting the numerator in order to cancel out the denominator. So here, since our numerator is 25 factorial and our denominator is 23 factorial, we want to rewrite that 25 factorial as 25 times 24 times 23 factorial using what we know about factorials here. Then since this gets divided by 23 factorial, that will end up canceling out and give me my final answer of 25 times 24, which is simply equal to 600.

So there are 600 different ways that the teacher could choose a line leader and a door holder from their class of 25 different students. Now looking at our second example here, we see that on our homework, there are 10 fill-in-the-blank questions and a word bank of 14 words. If we can only use each word once, how many possible ways could we answer these 10 questions? So again, the first thing that we want to do here is identify our values of n and r. So remember, n is always going to be your largest number, in this case, 14.

This represents my total. Then r is my smaller number that I'm choosing from my total. In this case, I'm choosing 10 words out of my total of 14. So 10 represents my value for r. Now writing this in our permutations notations, this is written as p14,10.

Now we're just left to plug everything into our formula here. So in this formula, we're going to take that 14 factorial on the top, our largest number. And then on the bottom, we have 14 minus 10 factorial. Now this simplifies to 14!−4! and again here we want to go ahead and rewrite the numerator to cancel the denominator. Now something that you might notice here is that we actually could just type this into our calculator if we know how to use the factorial button.

And there actually is a specific function on your calculator to do permutations. But if you're not quite sure how to use that, we can always work this out by hand. So let's continue here and rewrite our numerator to cancel our denominator. Now this one is going to get a bit long but that's okay so let's go ahead and start. Now on that numerator, I can simplify this to 14 times 13 times 12 all the way down to five, multiplying all of that along the way. Now if I type this in my calculator now, multiplying all of that, I'm going to get a value of 3,632,428,800. So these are all of the different ways that we could fill in these 10 questions. So it's safe to say that guessing is not going to be your best bet here. So now that we know what permutations are and how to calculate them, let's get some more practice.

Let me know if you have any questions.

A student formed a club at their school. They have 13 members, and need to elect a president, vice president, and treasurer. How many ways are there to fill these officer positions?

• A. $2197$
• B. $1716$
• C. $13$
• D. $6$

Answer: (B)

Emily is organizing her closet. She has 15 shirts left to hang but has space in one section for 6 shirts. How many ways could she hang shirts in that section?

• A. $3,603,600$
• B. $90$
• C. $9$
• D. $362,880$

Answer: (A)

Evaluate the given expression. $_9P_4$

• A. $24$
• B. $3,024$
• C. $15,120$
• D. $362,880$

Answer: (B)

Permutations of Non-Distinct Objects

Hey everyone! We now know how to find permutations when dealing with multiple different or distinct objects, like picking five different shirts to wear over five days or choosing six marbles out of a bag of six different colored marbles. Well, what if instead in that bag there were four blue marbles and two red marbles? If I chose this blue marble on the first try, wouldn't it be the same thing as if I were to choose this blue marble? Well, yes, it would be the same thing, which means that we're going to have a different number of outcomes here.

So when faced with multiple objects that are the same or non-distinct, we're going to have to go about it a little bit differently. Now, just saying that permutations of non-distinct objects sound really complicated but you don't have to worry because here we're just going to take our equation for permutations that we already know and just alter it slightly to account for these objects that are the same. And here I'm going to walk you through exactly what that change is and how we're going to use this equation. So let's go ahead and get started. When finding all of the outcomes of drawing six different marbles out of a bag of six different colored marbles, simply using our permutations equation, we're going to take our total and the factorial of that and divide it by our total minus the number of things that we're choosing and the factorial of that.

Now since the number of things we're choosing happens to be the same thing, I get 6!(6-6)6! using this formula here. And when looking at our permutations of non-distinct objects with our bag of four blue marbles and two red marbles, we're going to start out the same way, so still taking our total number of objects and the factorial of that on the top. Now here, since I still have a total of six, I get 6!4!×2! in that numerator. Now in the denominator is where we're going to account for objects that are the same. So we need to consider all of the different types of objects that we have.

Now in this case, we have blue marbles and we have red marbles, two different types of objects. So we want to look at the number of different types of objects that we have and take the factorial of that on the bottom. So since I have two different types of objects, I'm going to make sure and count those and take the factorial of the number of each of them on the bottom. So here since I have four blue marbles, I'm going to take four factorial and then I have two red marbles so I'm going to multiply that four factorial by two factorial and this is my final equation here and I would just solve that as we already know how. Now that we've seen that equation, let's go ahead and work through some examples to make this a bit more clear.

So looking at our first example here, we are asked how many different eight-digit codes can be made from five zeros and three ones. Now the first thing we want to do here as we did with permutations for distinct objects is still identify what our value of n and what our different values of r are. So here remember n is your highest number it's your total. So in this case, I have a total of eight digits, so this represents my value for n. Then since I have two distinct objects, I have zeros and I have ones, that means I have this value for r one, which is five, and then I have this value for r two, which is three.

Now it doesn't really matter which one is r one and r two so long as you're accounting for them both. So let's go ahead and write our permutations formula following this here. So we're going to again take our total on the top, our value for n, so 8!5!×3!. Now, since we have multiple factorials on the bottom, we are simply going to rewrite our numerator to cancel the highest factorial to simplify this in the most effective way.

So from here, we're going to want to cancel this five factorial. So let's rewrite eight factorial, our numerator, in order to cancel that. So I can rewrite eight factorial as eight times seven times six times five denominator I have five factorial and three factorial. Now those five factorials are going to cancel leaving me on the top with eight times seven times six and in my denominator with three factorial, which expanded out, is simply three times two times one. Now we could choose to cancel some more things here because you might see a couple of things that cancel, but I'm going to go ahead and just fully multiply this out in order to simplify from there.

So on the numerator, I end up getting 336 and then in the denominator, I get six. Now dividing this out, I end up with a final answer of 56 different eight-digit codes that contain five zeros and three ones. Let's take a look at our second example here. We're asked how many different ways there are to arrange the letters of the word banana. Now this might seem like a kind of strange question, but we can still attack it the same way.

Now the first thing again we want to do is identify our different values here. So we want to find n, and we want to find any values of r, r one, r two, and if I have more than that as well. So here, since we're looking at this word banana, we want to know how to arrange all of these different letters. So that means my value of n is going to be the total number of letters in this word. Now the word banana has six different letters, so that tells me that my value of n is equal to six.

Now we need to identify all of the different types of objects that we have here, in this case, any letters that are the same in this word. Now starting out with b, that's going to be my first type. There's only one b in the word banana, so that means my value for r one is simply going to be one. Then I get to my a, and there are actually three a's in the word banana, so that tells me that my r two is going to be three. Now finally, I have n here and there are two n's in the word banana, so r three is going to be two.

Now from here since we have our n and all of our values of r, we can go ahead and use our formula. So I'm going to take that n factorial on the top as I always do. So six factorial. And then I'm going to divide that by all of my different types of objects and the factorials of those. So I have one factorial times three factorial times two factorial.

And remember, we're going to rewrite the numerator in order to cancel the highest factorial in the denominator here. Now, my highest factorial here is this three factorial. So we're going to go ahead and rewrite that numerator to cancel that out. Now doing that, I get six times five times four times three factorial. And then in my denominator, I still have that one factorial times three factorial times two factorial.

Now those three factorials will, of course, cancel, leaving me with my numerator as six times five times four. And then in my denominator, I know that one factorial is just one, so it's not going to do anything there. And then two factorial is two times one. Now fully multiplying that out, that gives me 120 in my numerator divided by two in my denominator to give me a final answer of 60 different ways to arrange the letters in the word banana. Now that we've seen how to work with permutations of non-distinct objects, let's get some more practice.

Thanks for watching and I'll see you in the next one.

You want to arrange the books on your bookshelf by color. How many different ways could you arrange 12 books if 4 of them have a blue cover, 3 are yellow, and 5 are white?

• A. $120$
• B. $11,880$
• C. $27,720$
• D. $479,001,600$

Answer: (C)

How many ways are there to arrange the letters in the word CALCULUS?

• A. $40,320$
• B. $5,040$
• C. $720$
• D. $6$

Answer: (B)

Permutations vs. Combinations

Hey, everyone. Up to this point, we've been working with the permutations of different objects. So when considering the permutations of two letters from a, b, and c, I would consider a b, but I would also consider b a. These are two different permutations because they're in two different orders. But now we're going to look at a new way to organize objects called a combination.

Now distinguishing between permutations and combinations can be tricky, especially because these words are sometimes used interchangeably in everyday language. But here, I'm going to walk you through exactly what the mathematical difference is between these two things and how to distinguish between them by considering just one thing: whether or not the order of these objects matters. So let's go ahead and jump right in here. Now in working with permutations, we saw that they were a different way to arrange objects in a particular order. So we know that when working with permutations, the order definitely does matter.

So when writing all the permutations of two letters from a, b, and c, I would want to consider b, c, but also c, b, and a, c, but also c, a. Because these are in completely different orders, they represent different permutations. But when working with combinations, combinations are simply a way to group objects. So if I consider groups of two letters from a, b, and c, I might consider a b, b c, and a c, but nothing else because when working with a b, this is the same group of letters as if I were to have b a. So when working with combinations, the order does not matter at all.

Now this might seem a little abstract, so let's go ahead and look at some different examples and determine whether we're working with a permutation or a combination. So looking at this first example here, I'm told that an ice cream shop has 32 different flavors, and we need to pick two flavors to blend into a milkshake. Now we want to know how many possible ways we can select these flavors. Here we're just concerned with identifying whether this is a permutation or a combination, so we want to ask ourselves one question: Does the order of my objects matter?

Well, if I'm blending up a milkshake, I might want to consider the outcome. So if I pick chocolate and vanilla and I blend that up, wouldn't that be the same milkshake as if I picked vanilla and then chocolate? Well, it would, so my outcome would be no different even if I changed the order. So since the order does not affect the outcome, here I'm simply working with a combination. The order does not matter.

Let's look at our next example. Here we're asked how many ways a photographer could line up the members of a family of five. Well, if I were taking this photo, I might consider a couple of different things. I might consider the height of the people in it or the outfits that they are wearing. So if I put one person on the end in one of those photos but then I move them to the other side in another photo, this would be a completely different photo.

So if I change the order, that affects the outcome. So because the order affects the outcome, that tells me that order does matter and here we're working with a permutation. Let's look at one final example here. Here we're asked how many different teams of four people can be formed from a group of nine people. Well, let's say that I pick four people out of this group of nine.

Well, if I pick this person first instead, is that going to change what my team is? Well, it's not going to change it at all besides maybe starting a fight between teammates. So because the order does not have an impact on the outcome, here the order does not matter and we're working with a combination. So now that we know how to distinguish between permutations and combinations, let's keep going. Thanks for watching and I'll see you in the next one.

Combinations

Hey, everyone. When finding the number of permutations of two letters out of these three letters, knowing that the order matters, whenever I list these out, I end up with six permutations. But when considering the combinations of two letters out of these three letters, knowing that the order does not matter, I only get three combinations. Now just seeing that these give me different numbers, you might realize that we're going to have to calculate these two things differently. But don't worry.

We're not just going to have to learn a brand new equation for combinations because we're just going to take our permutations formula and alter it slightly in order to account for the fact that order doesn't matter in order to get our new combinations formula. So with that in mind, let's go ahead and get started. Now when working with our combinations, if we just plug in our values for n and r, we end up getting three factorial over one factorial, which is just equal to six. Now, for our combinations formula, it looks almost identical to our permutations formula, but now we're just dividing by r factorial. So you'll see that I have my permutations formula right here, and I'm just going to add this extra factor of r factorial on the bottom here.

So whenever I compute my combinations formula, plugging in my values for n and r, I end up getting three factorial over one factorial times two factorial, which gives me my answer of three. Now with this formula, remember that n is always going to be your largest number because this represents your total and r is the number of things you're choosing out of that total. Now one way that you could remember this formula, seeing that we have an n and then an n and then an r and an r, whenever you think of combinations, you can think n, n, r, r in order to remember the order that these variables come in. Now something else that might seem familiar here is this combinations notation looks similar to our permutations. Well, with the c with the n and the r as subscripts or c with n and r in parentheses.

Both of these can be read as the number of combinations of r things taken n at a time. Now that we know what our combinations formula is, let's get a bit more practice using it by working through some examples together. So looking at this first example, we're told that an ice cream shop has 32 flavors and we need to pick two flavors to blend into a milkshake. We want to know how many possible ways we could do this. Now remember this is a combinations problem because the order doesn't matter.

A chocolate and vanilla milkshake is the same thing as a vanilla and chocolate milkshake. Now the first thing we want to do here is identify our values for n and r. Remember that n is always going to be the largest number, our total. So here my value for n is 32. Then this two right here represents my value for r.

So plugging this into my combinations notation, I get 32C2 and then I'm just plugging these numbers into the formula. So I start with 32 factorial on the top, your largest number, and then I have 32 minus two factorial and don't forget that extra factor of r factorial on the bottom there. Now simplifying this a bit more, I have 32 factorial on the top divided by 30 factorial times two factorial. Now here, since we have multiple factorials in that denominator, remember we can rewrite our numerator in order to cancel the highest factorial in the denominator. So here we want to cancel this 30 factorial, so let's rewrite that numerator to cancel it.

So 32 factorial can be rewritten as 32 times 31 times 30 factorial, and then in my denominator, I have 30 factorial times two factorial. Now those 30 factorials will cancel leaving me on the top with 32 times 31 and then on the bottom expanding that two factorial that's just two times one. Now actually multiplying that out gives me nine ninety-two on the top divided by two on the bottom giving me a final answer of four hundred ninety-six different possible ways we can select our flavors here. Let's move on to our next example here.

Here we're asked how many different teams of four people can be formed from a group of nine people. So let's first identify our values of n and r. Now remember n is your highest number, so this nine right here, and then r is our other number, so four. Now writing this as a combination, we have 9C4, and we can plug in those values to our formula. So starting with n on the top, of course, we have nine factorial divided by nine minus four factorial times four factorial.

Now, simplifying this a bit further, we end up with nine factorial over five factorial times four factorial. Now, again, we want to cancel the highest factorial in the denominator here by rewriting that numerator. So here I'm going to rewrite my numerator as nine times eight times seven times six times five factorial, and then on the bottom, I have that five factorial and four factorial. But those five factorials are going to cancel.

So, on the top, I am left with nine times eight times seven times six, and on the bottom, I have that four factorial, which expanded out is four times three times two times one. Now multiplying that out I get a rather high number in my numerator here 3,024, and then on the bottom I get 24. Now actually dividing that gives me a final answer of 126 different teams of four people that can be formed from that nine. Now that we know how to calculate combinations, let's get some more practice. Thanks for watching and let me know if you have questions.

From a class of 28 students, in how many ways could a teacher select 4 students to lead the class discussion?

• A. $491,400$
• B. $24$
• C. $20,475$

Answer: (C)

Evaluate the given expression. $_{11}C_7$

• A. $330$
• B. $120$
• C. $5,040$
• D. $7,920$

Answer: (A)

Counting Example 1

Hey everyone. In this problem, we're told that whenever playing a particular lottery game, you must choose five numbers between one and forty and you win if those five numbers match those drawn in the lottery. So we want to find the probability that you will win if you purchase one lottery ticket and if you purchase 50 different ones. So let's go ahead and get started here. Now, in calculating probability, we have to take the number of outcomes that include our event and divide that by the number of total possible outcomes.

But here, there are a lot of possible outcomes because there are a lot of different ways that you could choose five numbers between one and forty. Now I don't want to have to write all of those out and figure them out manually, so we actually can use a tool that we've used in the past year, and that's our combinations formula. We can use this to calculate all of the possible combinations of five numbers between one and forty. So before we can calculate our probability, we first need to come up with the number of total possible outcomes using that combinations formula. So here, we see that our value for n is going to be equal to 40 because we're choosing out of 40 different numbers.

But out of those 40 numbers, we're only choosing five. So that represents our value for r. So here, I'm going to calculate C405 or the combinations of five objects out of 40 total possible objects. So here, plugging these values into my formula, we're going to go ahead and take 40!35!5!. Now, from here, we can simplify this to 40!35!5!.

Now from here, if you know how to use the factorial button on your calculator, you can go ahead and just type this in and go from there. But here, I'm going to walk you through how to do it by hand just in case you want to see that. So from here, we're going to rewrite that numerator in order to cancel out the highest factorial in our denominator, which in this case is 35 factorial. So doing that we can rewrite that numerator as 40 times 39 times 38 times 37 times 36 finally times 35 factorial.

Now all of this, of course, is divided by 35 factorial times five factorial, and that 35 factorial on the top and the bottom cancels just as we wanted it to. Now in my numerator, I'm left with this 40 times 39 times 38 times 37 times 36, and in my denominator, I'm left with five factorial. Now that five factorial can get expanded out into five times four times three times two times one. So in my numerator, I have some multiplication happening as well in my denominator. Now doing that multiplication and then dividing that numerator by my denominator I get a final answer of 658,008 total possible combinations here.

So this tells me the number of total possible outcomes when calculating my probability. Now we can go ahead and calculate our probability. So the probability of winning with one single lottery ticket, if I purchase one lottery ticket, there's only one possible way that I could win if my one lottery ticket matches what's drawn. So I'm going to take that one and divide it by my total, 658,008. Now this as a fraction might not mean much to you, but if we put this as a decimal, this gives me a value of 0.0000152, and this is our probability of winning with one single lottery ticket.

So our chances don't look too great here, but let's look if we were to buy 50 different lottery tickets. So here now the probability of winning there are now 50 different ways I could win if any one of my 50 purchased lottery tickets matches that that was drawn. So if I take that 50 and still divide it by that same total, 658,008, I get a decimal value of 0.000076. So a little bit higher possibility of winning the lottery here, but maybe not quite as high as we'd want to see. So now that we know how to find probabilities also using our combinations formula, thanks for watching, and I'll see you in the next one.

Here’s what students ask on this topic:

What is the difference between permutations and combinations?

Permutations and combinations are both methods of counting arrangements, but they differ in whether order matters. In permutations, the order of arrangement is crucial. For example, the arrangement ABC is different from BAC. The formula for permutations is given by $\frac{n!}{(n-r)!}$, where n is the total number of items, and r is the number of items being chosen. In combinations, the order does not matter. For instance, the group ABC is the same as BAC. The formula for combinations is $\frac{n!}{r!(n-r)!}$. Understanding these differences is crucial for solving problems in probability and statistics.

How do you calculate permutations of non-distinct objects?

To calculate permutations of non-distinct objects, you modify the standard permutation formula to account for repeated items. The formula is $\frac{n!}{r{!}_{1}r{!}_2...}$, where n is the total number of objects, and r₁, r₂, etc., are the counts of each type of indistinguishable object. For example, in the word 'banana', n = 6, with r₁ = 3 for 'a's, and r₂ = 2 for 'n's. The formula becomes $\frac{6!}{3!2!}$, simplifying to 60 different arrangements.

How do you determine if a problem involves permutations or combinations?

To determine if a problem involves permutations or combinations, ask whether the order of selection matters. If the order is important, such as arranging people in a line, it is a permutation problem. Use the formula $\frac{n!}{(n-r)!}$. If the order does not matter, like choosing a committee from a group, it is a combination problem. Use the formula $\frac{n!}{r!(n-r)!}$. This distinction is crucial for correctly solving problems in probability and statistics.

What is the formula for calculating combinations?

The formula for calculating combinations is used when the order of selection does not matter. It is given by $\frac{n!}{r!(n-r)!}$, where n is the total number of items, and r is the number of items being chosen. This formula accounts for the fact that different orders of the same group are considered the same. For example, choosing 2 flavors from 32 for a milkshake, where order doesn't matter, would use this formula to find the number of possible combinations.

How do you calculate permutations when order matters?

To calculate permutations when order matters, use the formula $\frac{n!}{(n-r)!}$, where n is the total number of items, and r is the number of items being arranged. This formula considers different orders as distinct arrangements. For example, arranging 3 out of 5 books on a shelf would use this formula to determine the number of possible sequences. Understanding this concept is essential for solving problems where the sequence of selection is important.