WEBVTT
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Find the number of ways to select three different letters from the set π, π, π, π, π, β, π, π.
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Well weβve got eight different letters that we can choose from, and we want to select three of them.
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Now given that theyβre all different letters, when I come to choose my first letter, Iβve got a choice of eight that I can choose from.
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So in the first box, Iβve got eight.
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Now when Iβve picked one letter, I canβt choose that one again because theyβve all got to be different letters, so Iβve now only got seven different ones to choose from.
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So on the second box, Iβve got a choice of seven letters.
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And having picked out those two letters, Iβve now only got six to choose from in the last box.
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Now each of those eight ways for the first box can be combined with each of those seven ways for the second box, and they can be combined with each of the six ways for the last box.
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So that means weβve got eight times seven times six, which is 336 different ways.
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But thatβs not quite the answer, Letβs say weβve picked π for the first box and π for the second box and π for the third box.
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We might have chosen the letters π and π in a different order, or maybe we picked π first, or even π.
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So within the 336 ways of organizing our three letters, chosen from those eight letters, each group of three has been represented in six different orders.
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So itβs the same group of three letters represented six different ways, so weβve kind of multiply counted some of our combinations.
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So given that Iβve not specified that Iβm interested in the order that the letters come out and Iβm gonna group all those six different combinations there as the same group of three letters, Iβm only going to have a sixth as many ways of picking out three letters from that group of eight.
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So the total number of ways is 336 divided by six, which is 56.
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And thatβs our answer.
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But before we finish, letβs just talk about some alternative notation.
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We have a general formula for this sort of problem.
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How many ways to choose π different objects from a set of π different objects?
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And we call that an π choose π type of problem.
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Now depending on where you live, youβll have seen one of these ways of expressing that particular formula: ππΆπ, πΆπ,π, πΆ(π,π), πΆππ, ππΆπ, or just π over π in kind of a vector format.
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But they all boil down to this basic formula: π factorial over π factorial times π minus π factorial.
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Now in our question we were choosing from eight different objects, so π was equal to eight.
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And we were selecting three of them, so π is equal to three.
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So our calculation becomes eight factorial over three factorial times eight minus three factorial.
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Now eight minus three is five, so that becomes five factorial on the denominator.
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Now if we think about what factorial means, so eight factorial means eight times seven times six times five times four times three times two times one; three factorial is three times two times one; and five factorial is five times four times three times two times one.
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Now we can do some cancelling; weβve got one on the top, one on the bottom; two on the top, two on the bottom; three on the top, three on the bottom; four on the top, four on the bottom; five on the top, and five on the bottom.
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And now we got back to eight times seven times six over three times two times one.
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And thatβs the 336 over six that we talked about when we did the problem the first way.
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So this method also gives us the same correct answer: 56.