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## Precalculus

### Course: Precalculus > Unit 8

Lesson 4: Combinations# Intro to combinations

CCSS.Math:

Sal introduces the basic idea of combinations.

## Want to join the conversation?

- Are permutations and combinations the same thing? I thought that in combinations you couldn't use the same people. Like in combinations we could do :CBA but we couldn't do BCA because they are the same people. And in permutations we could do :CBA and BCA because order didn't matter. I'm soooo confused!(34 votes)
- In Permutations the order matters. So ABC would be one permutation and ACB would be another, for example. In Combinations ABC is the same as ACB because you are combining the same letters (or people). Now, there are 6 (3 factorial) permutations of ABC. Therefore, to calculate the number of combinations of 3 people (or letters) from a set of six, you need to divide 6! by 3!. I think its best to write out the combinations and permutations like Sal does; that really helps me out.(55 votes)

- it says combinatations do not matter

however suppose there is a locker combination, the order matters so why is a combianation nmattering(7 votes)- You're talking about a permutation, even though in the real world people use the word combination (which is mathematically wrong).

Here's an easy way to remember:

- If a group consisting of Alice, Bob and Charlie has the same meaning as a group consisting of Charlie, Alice and Bob, you're talking about a**combination**.

- If a group consisting of Alice, then Bob and THEN Charlie is NOT the same meaning as a group consisting of Charlie, then Alice and THEN Bob, you're talking about a**permutation**.*Remember to upvote good questions and helpful answers*(107 votes)

- I don't understand, at4:50Salman says, to find the number of ways to arrange three people from the six, the equation is 120 (total number of permutations) divided by 6 (number of ways to arrange the letters in a set). Why wouldn't the equation to find all arrangements of three people be the same as finding the total number of possibilities in a set? (3 letters*3chairs = 9 different arrangements)(6 votes)
- There are 3 people who can sit in chair one. Then there are only two of those three left for chair two and then one for chair three. 3*2*1 equals 6(26 votes)

- Just being curious, is the word 'permutation' in any way related to mutation

(per-mutation)?(5 votes)- Yes it is. A "mutation" is a change, and the prefix per- means something like "very" or "thorough". So a permutation can be interpreted as a "thorough change".(17 votes)

- In How many ways can 5 letters be posted in 4 postboxes if each postboxes can contain any number of letters ?

Answer is 4^5 = 1024 but my question is why it is not 5^4? Please help, thank you..(3 votes)- Each of the 5 letters has 4 possibilities for where it can be, so the number of results is 4*4*4*4*4 or 4⁵(15 votes)

- How to remember the difference between combination and permutations?(4 votes)
- combinations:(C)arefree of the order.

permutations: needs order, very (P)ragmatic.

P is a pragmatic parent. C is a carefree child.

Saw that in some website months ago.....(9 votes)

- Why is 6 x 3 not right to find the number of combinations?(4 votes)
- Why is it FBC? What is FBC?

Why isn't it DBC?(2 votes)- FBC is a combination of 3 people: person F, person B, and person C.(4 votes)

- this video helped sm, thanks sal!(3 votes)
- At5:00, I don't understand why I would have to divide by 6. I just don't understand where the 6 comes from and why. Someone please help me explain?

Thanks(2 votes)- When dealing with
*permutations,*the order in which we pick the three people matters.

But when dealing with*combinations,*the order doesn't matter.

For example, 𝐴𝐵𝐶 and 𝐵𝐴𝐶 are different permutations, but not different combinations.

For each combination of three people, like 𝐴𝐵𝐶, we can construct six permutations. In this case they would be 𝐴𝐵𝐶, 𝐴𝐶𝐵, 𝐵𝐴𝐶, 𝐵𝐶𝐴, 𝐶𝐴𝐵 and 𝐶𝐵𝐴.

– – –

In the video, Sal finds that there are 120 possible permutations when choosing three people from a total of six people.

These 120 permutations can be divided into groups, such that each group consists of the permutations that represent the same combination.

Since we are choosing three people, each group would consist of 6 permutations.

Thereby there would be 120∕6 = 20 groups,

and because each group represents a unique combination, we can then conclude that the number of possible combinations when choosing three people from a total of six people is 120∕6 = 20.(3 votes)

## Video transcript

- So let's keep thinking
about different ways to sit multiple people in
the certain number of chairs. So let's say we have six people. We have person A, we have
person B, we have person C, person D, person E, and we have person F. So we have six people. And for the sake of this
video, we're going to say oh we want to figure
out all the scenarios, all the possibilities,
all the permutations, all the ways that we could
put them into three chairs. So there's chair number
one, chair number two and chair number three. This is all review. This is covered in the permutations video. But it'll be very instructive as we move into a new concept. So what are all the
permutations of putting six different people into three chairs? Well, like we've seen before, we can start with the first chair. And we can say look if no one's sat-- If we haven't seated anyone yet, how many different people could
we put in chair number one? Well there's six different
people who could be in chair number one. Let me get a different color. There are six people who
could be in chair number one. Six different scenarios for
who sits in chair number one. Now for each of those six scenarios, how many people, how many different people could sit in chair number two? Well each of those six
scenarios we've taken one of the six people to
sit in chair number one. So that means you have
five out of the six people left to sit in chair number two. Or another way to think about it is there's six scenarios of
someone in chair number one and for each of those six,
you have five scenarios for who's in chair number two. So you have a total of 30 scenarios where you have seated six
people in the first two chairs. And now if you want to say well what about for the three chairs? Well for each of these 30 scenarios, how many different people could you put in chair number three? Well you're still going
to have four people standing up not in chairs. So for each of these 30 scenarios, you have four people who you could put in chair number three. So your total number of scenarios, or your total number of permutations where we care who's sitting in which chair is six times five times four, which is equal to 120 permutations. Permutations. Now lemme, permutations. Now it's worth thinking about what permutations are counting. Now remember we care,
when we're talking about permutations, we care about who's sitting in which chair. So for example, for example
this is one permutation. And this would be counted
as another permutation. And this would be counted
as another permutation. This would be counted
as another permutation. So notice these are all
the same three people, but we're putting them
in different chairs. And this counted that. That's counted in this 120. I could keep going. We could have that, or we could have that. So our thinking in the permutation world. We would count all of these. Or we would count this as
six different permutations. These are going towards this 120. And of course we have other permutations where we would involve other people. Where we have, it could
be F, B, C, F, C, B, F, A, C, F, F. Actually let me do it this way. I'm going to be a little
bit more systematic. F, uh lemme do it. B, B, F, C, B, C, F. And obviously I could keep doing. I can do 120 of these. I'll do two more. You could have C, F, B. And then you could have C, B, F. So in the permutation world, these are, these are literally 12
of the 120 permutations. But what if we, what if all we cared about is the three people we're choosing to sit down, but we don't care in what order that they're sitting, or in which chair they're sitting. So in that world, these would all be one. This is all the same set of three people if we don't care which
chair they're sitting in. This would also be the
same set of three people. And so this question. If I have six people
sitting in three chairs, how many ways can I choose
three people out of the six where I don't care
which chair they sit on? And I encourage you to pause the video, and try to think of what that
number would actually be. Well a big clue was when we essentially wrote all of the permutations
where we've picked a group of three people. We see that there's six ways
of arranging the three people. When you pick a certain
group of three people, that turned into six permutations. And so if all you want to do is care about well how many different
ways are there to choose three from the six? You would take your whole permutations. You would take your
number of permutations. You would take your
number of permutations. And then you would divide
it by the number of ways to arrange three people. Number of ways to arrange,
arrange three people. And we see that you can
arrange three people, or even three letters. You can arrange it in six different ways. So this would be equal
to 120 divided by six, or this would be equal to 20. So there are 120 permutations here. If you said how many different
arrangements are there of taking six people and
putting them into three chairs? That's 120. But now we're asking another thing. We're saying if we start with 120 people, and we want to choose. Sorry if we're starting with six people and we want to figure out how many ways, how many combinations,
how many ways are there for us to choose three of them? Then we end up with 20 combinations. Combinations of people. This right over here, once again, this right over here is
just one combination. It's the combination, A, B, C. I don't care what order they sit in. I have chosen them. I have chosen these three of the six. This is a combination of people. I don't care about the order. This right over here
is another combination. It is F, C, and B. Once again I don't care about the order. I just care that I've
chosen these three people. So how many ways are there to choose three people out of six? It is 20. It's the total number of permutations. It's 120 divided by the number of ways you can arrange three people.