Combinations Introduction to combinations
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- In the last video we figured out how many different ways
- could 5 people sit in 3 chairs.
- So for example, if this is chair one, this is chair
- two, this is chair three.
- We said, well, in chair one we could put 5 people.
- No one's sitting down.
- Then there's only going to be 4 people left, so we could put 4
- different people in chair 2.
- And then there would be 3 people left that we could
- put in chair three.
- So the total permutations, the different ways that people
- could sit in the different yours if we cared about the
- order-- if we cared about which chair they were sitting down
- in-- it would be 5 times 4 times 3.
- And another way of thinking about that-- 5 times 4 times
- 3-- that's the same thing.
- That equals 5 times 4 times 3 times 2 times 1 over what?
- Over 2 times 1.
- And that's the same thing as 5 factorial over 2 factorial.
- Now where does this 2 come from?
- How is 2 related to the 5 and 3?
- What's the difference between the two?
- So this is the same thing, this is equal to 5 factorial
- over 5 minus 3 factorial.
- And that in general is how we figure out, how many
- permutations can 5 things place themselves or be
- placed into 3 positions?
- And the general formula is-- and we learned
- this in the last video.
- And I will switch colors.
- If we want to put n things into k positions, and k has to
- be less than or equal to n.
- Well actually, it doesn't have to.
- But for our purposes right now we'll assume it is because our
- formula might break down if it weren't.
- And that equals n factorial over n minus k factorial.
- I always find this harder to memorize than just
- thinking about the spot.
- Then just saying, oh well, you know, 5 people, 5 of
- the things could be here.
- And then once one thing is here, there's 4 possibilities
- left and then there's 3 possibilities left.
- The way I think of it I take the first k terms
- of the n factorial.
- Or in this case, I take the first three terms
- of the 5 factorial.
- 5 times 4 times 3.
- That's how I think of permutations.
- This is great if we cared-- let's say these are
- people A, B, C, D, E.
- So these are the 5 people that are going to sit in the chairs.
- This is great if we wanted to count the permutation ABC as
- different from the permutation ACB, as different from the
- permutation-- I don't know-- BAC as different from
- the permutation BCA.
- Because remember, when we did this we actually cared about
- where they're sitting.
- And in the previous video we double counted everything
- because it matters if person A is in seat one and
- person B is in seat two.
- And then if they switch we recounted, right?
- This is where they switch.
- But what if we didn't care about that?
- What if we didn't care who's in what seat?
- We just wanted to know how many different ways can
- the 5 people sit down?
- So we want to count all the situations where person A, B,
- and C are sitting down as essentially one situation.
- We don't care who's sitting in which chair.
- We just care that those are the 3 people sitting down.
- That's the set, the subset of the people sitting down.
- And so the question then becomes not how many different
- permutations or how many different ways can the people
- sit down, the question becomes, how many subsets of 3 can
- we take out of a set of 5?
- And I know I'm kind of jumping up around a little bit, but
- that's essentially what a combination is.
- A combination is a permutation where you don't care
- about the order.
- So how do we figure it out?
- Well, when we figured out the permutations using this formula
- we counted-- for example, we counted ABC, ACB, BAC,
- BCA, and let's see.
- There should be two more permutations.
- CAB and CBA.
- We counted all 6 of these as different permutations.
- But in our combinations we're going to want to-- this is all
- essentially the same combination because we don't
- care about the order.
- So for any 3 different people that are in these seats,
- there's actually going to be 6 permutations that we're
- counting when we do the permutations.
- So if we want the combinations we'll just divide by the number
- of ways we can rearrange 3 people into 3 seats.
- That's essentially what we did here.
- So how many different ways can you arrange 3
- people into 3 seats?
- Well, this is kind of another permutation problem.
- The first seat you could put 3 different people, the second
- seat you put 2 different people, and the last seat--
- well, there's only 1 person left.
- So it equals 3 factorial, which is equal to 6.
- This is equal to 3 factorial, which is equal to 6.
- Hope I'm not confusing.
- What I'm just trying to say is when you did a permutation we
- counted all of the different orders of how people could
- arrange themselves.
- And what I'm saying now is well, how many different ways
- can people arrange themselves?
- Well, it's going to be the number of places factorial
- because if we have 3 people in 3 spots or let's say,
- 4 people in 4 spots.
- The fourth spot can have 4 people, the second spot can
- have 3, and so forth, the third spot could have 2, and the
- last spot will only have 1.
- So it's the number of spots factorial is how many
- permutations we're counting.
- When we have just the same people, they're just playing
- musical chairs in the exact same seats.
- So on order to figure out the combination, so if we wanted to
- say how many people-- let's say if we had 5 people.
- How many different groups of 3 can be seated?
- And we don't want to double.
- We don't want to more than double count.
- I don't know what the word is for counting
- something six times.
- Well, it's just going to be the same thing as the permutation
- divided by all of the extra counting we did.
- We'll just divide by the number of ways that 3 people can
- arrange themselves in 3 seats.
- And that's 3 factorial.
- And I hope I'm making sense.
- Maybe I'll do a couple more examples in other videos.
- And definitely request it if you think this
- is extra confusing.
- So in general, if we say, what are the different ways that
- n things can be chosen?
- Or the number of combinations that n things can be chosen
- into sets of r, where r is less than or equal to n.
- It's equal to the number of permutations you could create
- of putting n things into r spots, divided by r factorial.
- We're going to divide by the number of ways the r spots
- themselves could be rearranged because we don't want to
- count those as extra.
- And so if we go back to this formula up here, well, this
- was a k, but now we're saying it's an r.
- This is the same thing as-- so the permutations was
- n factorial over n minus r factorial.
- And now we're dividing everything by r factorial.
- So that equals-- let me just write this.
- This is often written as n choose r.
- Another way it's written is n choose r.
- This is called the binomial coefficient and we'll do a
- whole series of modules on that as well because this actually
- shows up in polynomial expansion when you take
- polynomials to powers.
- But this is equal to n factorial over r factorial
- divided by n minus r factorial.
- You could memorize this.
- You know, it's useful if you want to do things
- quickly on tests.
- But it's very important to think about where it came from.
- The n factorial over the n minus r factorial-- this
- is just the permutation.
- And what is that?
- Well, that was just the first r-- I guess you could call
- it the first r factors.
- The r largest factors of n factorial.
- That's all that is.
- And then when we do the combinations we divided by r
- factorial because we want to divide it by all the different
- arrangements that the people could seat themselves
- in r seats.
- Or, the balls could be placed in r cups.
- So in this situation if we want to know how many different
- groups of 3 can be selected from 5 people or from 5
- letters, it's going to be 5 factorial over 3 factorial
- times 5 minus 3 factorial.
- And that's 5 times 3 times 2 times 1, over-- 3
- factorial is just 6.
- We'll put that aside for a second.
- Divided by-- this is 3 factorial.
- 2 times 1.
- So notice, this is a permutation part right here.
- This term just gets rid of the two lowest factors.
- You get 5 times 3.
- Oh, sorry there's a 4.
- 5 times 4 times 3, which is the number of permutations.
- And then we divide by 6 because we get 6 permutations for
- really every combination.
- Maybe that confused you.
- But anyway, so we get 5 times 4 times 3 divided by 6.
- And that's what?
- 5 times 12 divided by 6, which is equal to 5 times 2.
- There's 10 possible ways that we can take sets of 3
- from a group of 5 things.
- See you in the next video.
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