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## Precalculus

# Intro to combinations

CCSS.Math:

Sal introduces the basic idea of combinations.

## Want to join the conversation?

- How to remember the difference between combination and permutations?(2 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.....(8 votes)

- what is the actual difference between permutations and combinations?(0 votes)
- something like ordered pair and unordered pair of a writing a composite number as a product of 2 natural nos.(0 votes)

- Why is it FBC? What is FBC?

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

- Anthony, William, Christopher and Mathew are sharing 4 hats: red, green blue and gold. If Anthony doesn't wear red, william doesnt wear green,Christopher doesn't wear blue and Matthew doesnt wear gold. how many different ways can they share the four hats?(2 votes)
- If everyone picked a hat at random, there would be 4! = 24 ways to divide the hats.

That leaves us with five scenarios:

1. No one gets a hat they will wear.

2. Three of them get a hat they will wear and one of them doesn't.

3. Two of them get a hat they will wear and two don't.

4. One of them gets a hat they will wear and three don't.

5. All of them get a hat they will wear.

Scenario 1.

This one is easy. There is only 1 way that everyone gets the one particular color they won't wear.

Scenario 2.

This one is also easy. If three of them get a hat they won't wear, then the fourth person automatically also gets a hat they won't wear.

For example, if Anthony gets the red hat, William gets the green hat, and Cristopher gets the blue hat, then there's only the gold hat left for Matthew.

So this scenario can't happen.

Scenario 3.

This one is a little trickier. First of all, there are 4!∕(2!⋅2!) = 6 ways to choose which two people get a hat they won't wear.

However, regardless of which two people we pick, the other two only have one way to divide the remaining hats.

For example if Anthony gets the red hat and William gets the green hat, then Cristopher must get the gold hat and Matthew the blue hat.

So that leaves us with a total of 6 ways.

Scenario 4.

This is where it gets tough.

Obviously there are 4 ways to choose which one gets a hat they won't wear.

Imagine that Anthony gets the red hat.

Then William gets either the blue or the gold hat.

If he gets the blue hat, then Matthew must get the green hat and Cristopher the gold hat.

If William gets the gold hat, then Cristopher must get the green hat and Matthew the blue hat.

So, in total there are 4⋅2 = 8 ways for this scenario to happen.

So, there is 1 way for scenario 1 to happen,

0 ways for scenario 2,

6 ways for scenario 3,

and 8 ways for scenario 4.

That leaves us with 24 − (1 + 0 + 6 + 8) = 9 ways for scenario 5, that everyone gets a hat they can wear.(4 votes)

- Why are we multiplying in permutation(3 votes)
- Use the example in the video. Like for the first chair, there are 6 possibility right? In the second chair, we have five people left, and each person will have 6 ways to sit with the previous person. Therefore, the probability for these people sitting in certain ways will be

. So that's the reason we do mutiply in permutations.`5*6=30`

(1 vote)

- @1:26how do you know to multiply? That is what doesn't seem so intuitive to me?(2 votes)
- Hi, I have a question.

'three girls of a group of eight are to be chosen. in how many ways can this be done?'

would the correct answer be 336??(2 votes)- The first girl can be any one of the eight girls so there are
**8**possibilities.

The second girl can be any one of the remaining seven girls so there are**7**possibilities.

The third girl can be any one of the remaining six girls so the number of possibilities is**6**.

Since all three girls are selected, multiplication will be applicable and all the separate possibilities will be multiplied. So 8*7*6=336.

*So yes, 336 is right,but only if the order of the girls (that are selected) matters*. If order in which the girls are selected does not matter then we divide our answer by the number of ways three girls can be arranged->3!, since we want to eliminate the possibilities in which the same three girls have been selected but are in different order.In that case the answer is 336/3!=336/6=56.The question should be more specific.(2 votes)

- what is the difference between permutation and combination?(2 votes)
- It's easier to understand when there is an example.

1) Permutation.

- This is used when you're lining up students. For example, there are students A, B, C, D, and E. When you are lining them up, the order matters.

2) Combination.

- This is used when you're making a group. However, the order does not matter because they're just in a group. If there are A, B, and C is in a group and B, C, and A is in a group, they're the same thing.

Hope this helps! If you have any questions or need help, please ask! :)(2 votes)

- If you are finding the combination for 5 things that are the same, like five paper clips. The order doesn't matter but how do you find the different ways to arrange 1-5 in order to find the combination? Or is there a way to find combination without find the ways to arrange thing?

(I have 2,000 objects and I'm trying to pick 5 of the same objects, that's my problem)(2 votes)- Each combination of 5 distinct objects can be arranged in

5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120 ways (permutations).

Picking 5 objects from 2,000 can be done in

2,000 ∙ 1,999 ∙ 1,998 ∙ 1,997 ∙ 1,996

≈ 31,840 trillion ways

Dividing that by 120 gives us about 265 trillion combinations.(2 votes)

- what is the main difference between permutations and combinations?they are almost the same thing in words but are different while calculating?(1 vote)
- If order doesn't matter, it's a combination.

If order does matter, it's a permutation,

Example:

The combination {𝐴, 𝐵, 𝐶} is equivalent to all of the possible permutations: {𝐴, 𝐵, 𝐶}, {𝐴, 𝐶, 𝐵}, {𝐵, 𝐴, 𝐶}, {𝐵, 𝐶, 𝐴}, {𝐶, 𝐴, 𝐵} and {𝐶, 𝐵, 𝐴}.(4 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.