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Current time:0:00Total duration:7:29

CCSS.Math:

let's say that there are four people in a room and you're probably tired of me naming the people with letters but I'm going to continue doing that so the four people in the room our persons a or people a b c and d and they are all told look you don't know each other so I want you to all meet each other so you need to shake the hand exactly once of every other person in the room so that you all meet so my question to you is if each of these persons each of these people need to shake the hand of every other person exactly once how many handshakes are going to occur the number of handshakes that are going to occur so like always pause the video and and see if you can make sense of this all right I'm assuming you've had a go at it so one way to think about it is okay if you say there's a handshake well a handshake has to two people are party to a handshake we're not talking about some new three person handshake or four person handshake we're just talking about the traditional two people shake their right hands and so there's one person there's another person that's party to and it's you say okay there's four possibilities of one party and if we assume people aren't shaking their own hands which we are assuming or they're always going to shake someone else's hand then for the other party there's only three other prop for each of these four possibilities who's this party there's three possibilities who's the other party and so you might say that there's four times three handshakes since there's four times three I guess you could say possible handshakes and what I'd like you to do is think a little bit about whether this is right whether there would actually be twelve handshakes well you might have thought about it you might say well you know this four times three this would count this is actually counting the permutations this is counting how many ways can you permute four people into two buckets the the two buckets of hand shakers where you care about which bucket they are and whether they're their hand shaker number one or hand shaker number two this would count this would count a being the number one hand shaker and B being the number two hand shaker as being different than B being the number one hand shaker and a being the number two hand shaker but we don't want both of these things to occur we don't want you know a to shake B's hand we're you know a is facing north and B is facing south and then another time we need to shake hands again where now B is facing north and a is facing south so we don't have to do it once these are actually the same thing so no reason for both of these to occur so we are going to be double counting so what we really want to do is think about combinations one way to think about it is you have four people in a world of four people or a pool of four people how many ways can you choose to how many ways to choose to choose two because that's what we're doing each handshake is just really a selection of two of these people and so we want to say how many ways can we select two people so that we have a different each combination each of these ways to select two people should have a different combination of people in it if two of them had this do the same kind of a b and b a these are the same combination and so this is really a combinations problem this is really equivalent to saying how many ways are there to choose two people from a pool of 4 or 4 choose 2 or 4 choose 2 and so this is going to be well how many ways are there where how many ways are there to permute 4 people into 3 spots which is going to be 4 times 3 which we just figured out right over there which is 12 actually wanted to do in that green color so you see where that came from so 4 times 3 and then you're going to divide that by the number of ways you would arrange to you can arrange two people well you can arrange two people in two different ways ones on the left ones on the right or the other ones on the left and the other ones on the right or you could also view that as 2 factorial which is also equal to 2 so we could write this down as 2 so this is the number of ways number of ways to arrange two people to arrange two people two people two people and this up here this up here is let me just in a new color this up here that's the permutations that's the ways number of permutations if you take two people from a pool of four so here you would care about order and so one way to think about it this too is correcting for this double counting year and if you want to apply the formula you could I just kind of reasoned through it again I mean you could literally say okay four times three is 12 we're double counting because there's two ways to arrange two people so you just divided it by two so you just divide by two and then you are going to be left with six you could think of it in terms of this or you could just apply the formula you could just say hey look for choose two or the number of combinations of selecting two from a group of four this is going to be 4 factorial over 2 factorial times 4 minus 2 4 minus 2 factorial and let me make this color different just so you can keep track of how I'm at least applying this and so what is this going to be this is going to be 4 times 3 times 2 times 1 over 2 times 1 times times this right over here is 2 times 1 so that will cancel with that 4 divided by 2 is 2 2 times 3 divided by 1 is equal to 6 and just really hit the point home let's actually draw it out let's draw it out so a could shake b's hand a could shake c's hand a could shake d's hand or b could shake and let me just do what we calculated first a 12 we could say B could shake a's hand b could shake c's hand B could shake whoops B could shake Dee's hand and you can say C could shake a's hands c could shake b's hand c could shake succeed could shake d's hand and you could say d could shake a's hand d could shake b's hand d could shake c's hand and this is 12 right over here and this is the permutations if d shaking c's hand is actually different than see taking shaking Dee's hand then we would count well but we just wanted to say well how many ways they just have to meet each other once and so we're double counting so so a B is the same thing as BA AC is the same thing as CA ad is the same thing as DA BC is the same thing as CB BD is the same thing as DB CD is the same thing as DC and so we'd be left with if we correct for the double counting we're left with one two three four five six combinations six possible ways of choosing two from a pool of four especially when you don't care about the order in which you choose them