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Introduction to combinations | Probability and Statistics | Khan Academy

Introduction to combinations | Probability and Statistics | Khan Academy


– 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.

Comments (41)

  1. So, is a combination lock misnamed? Is the true name of a combination lock a permutation lock?

  2. How to go about deriving the formula number of ways of choosing r objects from p objects of one type, q objects second type and so on

  3. He's used all three by now "have sitted", "have sat" and "have sitten".

  4. what about 5 in 4?

  5. How would you go about setting up a problem for seating these people (A – F) for more than 6 seats? Or is this addressed in a separate video?

  6. why is the first question a permutation not a combination since order of the seating doesnt matter?

  7. so what is the number of permutations if the number of chairs is 60 but the number of people is 5? Wouldn't the answer be negative according to the formula?

  8. Thanks a lot sir you clear out my confusion

  9. who else watched this inside school

  10. This is better than your old combinations video. Thank you! 🙂

  11. How many combinations of 6 digits are there if the pool of digits are 1-69 and a second pool 1-26. Conditions only allow digits 1-26 can be chosen twice, but only once per pool for each combination?

  12. thanks for not making me feel stupid anymore

  13. Great explanation ????

  14. I came for combinations not permutations

  15. 3:20 You could have KFC

  16. Well, you have actually not explained how to calculate combinations in the first place! You explained how to calculate permutations but how on earth should I know how to calculate combinations if I do not do permutations first?

  17. why dont i get it… what why is permutation related to combination at all…

  18. Answer should be 10 right. Because when we split 6 people into group of 3 then 6C3 = 20 but in that 20 , 10 will be the duplicates of other 10. So it should be 10. If it we are not splitting them into group of 3 then 6CK ( K = 3 ) is correct .

  19. Well explained! Thanks! All this while I didn't know the concept of combinations

  20. Imagine going to sleep every night , resting and knowing you have helped so many people out there to learn! Hats off to you man

  21. Why is this soooo hard

  22. Leave to a college course to complicate counting. lol

  23. You shouldve known that people who came here is for the
    2* in terms of c(n,r)

  24. i still dont get what the word permutation means

  25. made it simple thanks

  26. Our teacher: we name the people Person A, Person A sub 1, Person A sub 1 sub 1, Person A sub 1 sub 1 sub 1, Person A sub 1 sub 1 sub 1 sub 1, and Person A sub 1 sub 1 sub 1 sub 1 sub 1
    Me: can't you just say Person A, Person B, Person C, Person D, Person E, and Person F?
    Teacher: Shut up idiot & takes out stick
    Me: Takes phone and is about to call my parents
    Teacher: Ok let's go with your idea
    ?

  27. How come when u flip a coin 3 times the choose value for 2 heads is 3 ? HHT, HTH , THH ? Isn’t that a permutation instead of a combination yet the binomial coefficient uses a combination?

  28. 3:45 was a legendary aha moment for me!

  29. Waw noice tutoreal

  30. How could we possibly know the number of ways to arrange 3 people (i.e 6) without having to write out all the possible ways? Seems very inefficient to think out ABC, BAC, CAB, etc… especially if this number was much larger than just 3?
    Thanks!!

  31. This video Help me lot to understand the concept about combination thnx ?

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