Video Transcript
In part one, we looked at how to
count up the number of ways to arrange objects and use the factorial notation to
produce a formula for this. We also looked at counting how many
ways there are to pick a certain number of objects from a set and came up with the
permutations formula, 𝑛 𝑃 𝑟. Now we’re gonna use these things to
count combinations of objects where their order isn’t relevant and we’re gonna use
the combinations formula, 𝑛 𝐶 𝑟 in a few examples.
So remember, we learned last time
that 𝑛 factorial means 𝑛 times 𝑛 minus one times 𝑛 minus two, and so on and so
on, times three, times two, all the way down to times one. So, for example, five factorial
means five times four times three times two times one. And we also learned the formula 𝑛
𝑃 𝑟, permutations formula, is equal to 𝑛 factorial over 𝑛 minus 𝑟 factorial, so
we’re picking 𝑟 objects from a set of 𝑛 objects. Now we did say that zero factorial
is equal to one. So if 𝑛 and 𝑟 turn out to be
equal, it would mean that with zero factorial on the bottom, we’d just make that
equal to one. Well an example of that is in the
question how many ways are there to pick three letters from the letters 𝐴, 𝐵, 𝐶,
𝐷.
So in this case 𝑟 would be equal
to three cause that’s the number of letters we’re choosing, and 𝑛 would be equal to
four cause that’s the size of the set of letters that we’re choosing from. So 𝑟 would be three because that
is the number of letters we’re choosing, and 𝑛 would be equal to four because
that’s the size of the set of letters that we’re choosing from. So the answer to the question would
be four 𝑃 three, so that’s four factorial over four minus three factorial, which of
course is four factorial over one factorial. And when we worked that out, we get
an answer of twenty-four. So the important thing about this
is that the order of the letters was quite important because that twenty-four
combinations there includes all of these ones here. And these are basically all
variations on 𝐴, 𝐵, and 𝐶, so just add the letters 𝐴, 𝐵, and 𝐶 in different
orders. If it doesn’t matter to you what
order they’re in, effectively we can count all of those different combinations as
one choice, just one option. And it’s that process that we’re
going to look at in this video.
So, as we’ve just seen, 𝑛 𝑃 𝑟
tells us how many permutations there are for picking 𝑟 objects from a set of 𝑛
objects. And again, as we’ve just seen, each
group of 𝑟 letters — so we have three letters in that example 𝐴, 𝐵, and, 𝐶 — is
written 𝑟 factorial different ways. So three factorial different
ways. That’s three times two times one is
six different ways in the big list. So that big list contains a lot of
repeats if you’re not really interested in what order they appear in. So we’re gonna look at how to work
out how many different combinations we get when choosing 𝑟 objects from set of 𝑛
objects if we discard all the different rearrangements of the same letter. So 𝑛 𝑃 𝑟, as we’ve said, told us
how many unique permutations we had, and that list included 𝑟 factorial repeats of
groups of three letters. So if we just want to know for
example how many groups of three letters, or 𝑟 letters, that we’ve got in a final
answer, then what we need to do is take the answer 𝑛 𝑃 𝑟 and divide it by 𝑟
factorial. So there we are, 𝑛 𝐶 𝑟 is 𝑛 𝑃
𝑟 divided by 𝑟 factorial. So the formula for 𝑛 𝑃 𝑟,
remember, was 𝑛 factorial over 𝑛 minus 𝑟 factorial. If we divide that by 𝑟 factorial,
then it’s the same as multiplying by one over 𝑟 factorial and that gives us this
formula here, 𝑛 factorial over 𝑟 factorial 𝑛 minus 𝑟 factorial. So remember, 𝑛 𝑃 𝑟 counts the
permutations where different orders of the same group of letters count separately,
so we’ve kind of kept double, triple, quadruple counting things and so on. And 𝑛 𝐶 𝑟 counts the
combinations where different orderings of the group are combined together. So in our last example, all the
𝐴𝐵𝐶, 𝐴𝐶𝐵, 𝐵𝐴𝐶, and so on, that would count as multiple permutations in the
𝑛 𝑃 𝑟 formula, but we would count all six of those variations just as one
combination in the 𝑛 𝐶 𝑟 formula because they are all the same group of three
letters.
Right, let’s have a look at a
couple of examples then.
So if we’ve got ten discs labelled
𝐴 up to 𝐽, we take three discs at random and set them down in the order that they
were selected. How many permutations are there for
this? So 𝑛 would be ten in this case
because we’ve got a set of ten letters to choose from. And 𝑟 will be three because we’re
picking three from that set. Now we’re setting them down in
order, so 𝐴𝐵𝐶 for example would not be equivalent to 𝐶𝐵𝐴 and so on, so we want
to use the 𝑛 𝑃 𝑟 formula.
And substituting in those values
for 𝑟 and 𝑛, we’ve got ten 𝑃 three, so that’s ten factorial over ten minus three
factorial. And that gives us ten factorial
over seven factorial. Now, obviously, we could just put
that in our calculator and get our answer, but I’m just gonna write that out in full
for a moment. And when we do that, we can see
that we can do quite a lot of cancelling. So seven divided by seven is one,
six divided by six is one, five divided by five is one, four divided by four is one,
three divided by three is one, and so on. All of those cancel out, so we’re
ending up with just ten times nine times eight over one or just ten times nine times
eight. So that gives us seven hundred and
twenty different permutations that we can get. So that’s our answer. Now, just to sort of highlight the
difference between 𝑛 𝑃 𝑟 and 𝑛 𝐶 𝑟, remember that seven hundred and twenty
includes for every set of three letters like 𝐴𝐵𝐶; we’ve got six variations on
that in there. And if we’re prepared to say that
they’re all equivalent cause they’re just combinations of the letters 𝐴, 𝐵, and
𝐶, we can divide that seven hundred and twenty by six, in this case, three
factorial because we had 𝑟 was equal to three to work out the number of
combinations. So there’s the calculation then for
𝑛 𝐶 𝑟. If we were looking for how many
clusters of three letters, the answer would only be a hundred and twenty. But the question specifically asked
for permutations and it said that the order was important, so we settled down in the
order that they were selected. So this is the correct answer.
Just make sure that, on the page,
we’ve got the-the proper correct answer highlighted, and it’s obvious that this
other working out wasn’t just another guess of what the answer might be.
Okay, let’s look at this small
essay for our next question.
In a lottery, we choose five
letters of the alphabet. During the draw a machine loads
twenty-six balls, each way- each one has one of the letters 𝐴 to 𝑍 on them. It shuffles them and then it lets
out five. It doesn’t matter what order they
were drawn in if our five chosen letters match the machines, then we win! How many combinations of five
letters are there to choose from? So in this case the order doesn’t
matter, so we’re gonna be using the 𝑛 𝐶 𝑟 combination. And there are twenty-six letters of
the alphabet to choose from, so 𝑛 is equal to twenty-six. And we’re choosing five of those
letters, so 𝑟 is equal to five. So we’re choosing five, 𝑟 equals
five, from twenty-six, 𝑛 equals twenty-six. And it doesn’t matter what order
they were drawn in, so we’re using the 𝑛 𝐶 𝑟 formula which is twenty-six 𝐶 five
when we plug those values in for 𝑛 and 𝑟. So that gives us twenty-six
factorial over five factorial twenty-six minus five factorial, which is twenty-six
factorial over five factorial twenty-one factorial. Now again we can just plug this
number into a calculator and get our answer, but I’m just gonna take a little extra
step here and just kind of split this out slightly. Sometimes the numbers that you get
are too big to work on a calculator; but by spotting some things that you can
cancel, you can actually generate numbers which are small enough that your
calculator can still handle. And that’s, well, that’s not the
case here. We can, you know, this-this number
would work on a calculator, but let’s see the technique anyway. So haven’t quite got room to write
out twenty-six factorial in full but I know it’s twenty-six times twenty-five times
twenty-four times twenty-three times twenty-two times twenty-one, and so on and so
on. But of course this bit here,
twenty-one times all the numbers down to one, is just twenty-one factorial. So we’ve got this stuff here, times
twenty-one factorial on there-on the top and on the denominator. We’ve got a twenty-one factorial,
so in fact these twenty-one factorials will cancel out. So we’ve got this divided by five
factorial. So sixty-five thousand seven
hundred and eighty, that’s the number of combinations that are there. So if this whole thing was done at
random, that’s the number of different ways there are of choosing five letters, so
we’ve got one in sixty-five thousand seven hundred and eighty chance of winning this
particular lottery. Now as a bit of an aside, if it did
matter, if we had to match the order in which the letters came out as well, we
would’ve used the 𝑛 𝑃 𝑟 formula. And of course we’ve got groups of
five, so that’s gonna five factorial sort of different combinations of each grouping
of five letters, so the answer is gonna be five factorial times bigger than that:
seven million eight hundred ninety-three thousand six hundred. So actually you’ve got a far
smaller chance of winning that particular lottery if it does matter what order the
balls came out in, but of course that’s not the answer we were looking for in this
particular case.
Okay, finally then, let’s just use
our combination counting techniques to calculate some probabilities. So we’ve got a question here.
A bag contains five chocolate and
fifteen strawberry sweets. If someone chooses a sweet at
random, find the probability that they pick a chocolate one; and if two people
choose a sweet, each at random, find the probability that they both get
chocolate. So really we’ve got two questions,
let’s call them 𝐴 and 𝐵. Let’s look at the first one
first. Someone chooses a sweet at random,
find the probability it’s chocolate. Well, this is pretty
straightforward. The probability you get a chocolate
sweet is just the number of ways of getting a chocolate sweet divided by the number
of ways of choosing a sweet of any sort. And of course there are five
chocolate sweets and then there are fiv- fifteen strawberry sweets, so it’s five
divided by five plus fifteen, so that’s five over twenty. So our answer, the probability of
getting a chocolate sweet, is just five over twenty. Okay, let’s look at the second
question. Now we’re going to look at two
different methods of doing this question. The first method, which is maybe
the one that you’ve already thought of, we’ve got two events. So first of all person one gets a
chocolate sweet and then person two gets a chocolate sweet. So for those two things to happen
together, we have to have one and the other, we’re going to multiply those
probabilities together. So we’ve just worked out that the
probability of one person getting a chocolate sweet is five over twenty. And of course if they’ve picked a
chocolate sweet, one, there’s just gonna be four chocolate sweets left, and the
other thing, there’ll now only be nineteen sweets in total for them to choose
from. So we’re gonna end up with
five-twentieths as the probability of the first person getting chocolate and
four-ninetieths as a probability of the second person getting a chocolate. Multiply those two together, do a
bit of cancelling, and we get one over nineteen. Again, in probability, you’re not
gonna get penalised if you don’t cancel those numbers down. But in this case, this three
hundred and eighty, has may be lost a little bit of its meaning, so it’s probably
more sensible to cancel down to those nice and simple numbers, one over
nineteen. So method two then, the probability
they both get chocolate is the number of ways that two people can choose chocolate
divided by the number of ways that two people can just choose any sweet. So the number of ways that two
people can choose chocolate is five 𝐶 two which would be choosing from five
chocolates, and there’s two of them, so five 𝐶 two is that. And the number of ways that they
can choose a sweet, but we’ve got twenty sweets in total and there’s two of them
choosing, so the total number of combinations of ways they can choose is twenty 𝐶
two. And if you do this on your
calculator, five 𝐶 two turns out to be ten and twenty 𝐶 two turns out to be a
hundred and ninety, which reassuringly when we cancel down we get the same answer as
we did the other way, one over nineteen. So there’s a one in nineteen chance
that both of them will get chocolate.
So just to summarise what we’ve
learned over the course of these two videos, we’ve learned about 𝑛 factorial, which
has got different ways of expressing it, but it’s 𝑛 times the number one smaller
times the number one smaller, and so on and so on, until we get down to one. And remember, we’ve got that
special definition of zero factorial is actually equal to one; so not entirely
intuitive, but you have to remember that. We’ve learned about permutations,
so 𝑛 𝑃 𝑟 and the various different ways of notating that in different regions, 𝑛
factorial over 𝑛 minus 𝑟 factorial. Now this counts every different
rearrangement of the same groupings as letters as individual cases, so we get quite
a large number in that case. But if we want to merge together
all different arrangement of the same letters that we’ve chosen and get a smaller
answer, we have to divide that by 𝑟 factorial. And because we’ve got all those
repeats of the same combinations of those letters or objects, we get combinations;
we get the 𝑛 𝐶 𝑟 formula. And there’s different ways again of
notating that, and that gives us the formula 𝑛 factorial over 𝑟 factorial 𝑛 minus
𝑟 factorial.
