### Video Transcript

Find the number of ways to select
three different letters from the set π, π, π, π, π, β, π, π.

Well weβve got eight different
letters that we can choose from, and we want to select three of them.

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. So in the first box, Iβve got
eight. 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. So on the second box, Iβve got a
choice of seven letters. And having picked out those two
letters, Iβve now only got six to choose from in the last box. 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.

So that means weβve got eight times
seven times six, which is 336 different ways. 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. We might have chosen the letters π
and π in a different order, or maybe we picked π first, or even π.

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. So itβs the same group of three
letters represented six different ways, so weβve kind of multiply counted some of
our combinations.

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. So the total number of ways is 336
divided by six, which is 56.

And thatβs our answer. But before we finish, letβs just
talk about some alternative notation. We have a general formula for this
sort of problem. How many ways to choose π
different objects from a set of π different objects? And we call that an π choose π
type of problem.

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. But they all boil down to this
basic formula: π factorial over π factorial times π minus π factorial.

Now in our question we were
choosing from eight different objects, so π was equal to eight. And we were selecting three of
them, so π is equal to three.

So our calculation becomes eight
factorial over three factorial times eight minus three factorial. Now eight minus three is five, so
that becomes five factorial on the denominator. 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.

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. And now we got back to eight times
seven times six over three times two times one.

And thatβs the 336 over six that we
talked about when we did the problem the first way. So this method also gives us the
same correct answer: 56.