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Find how to use the permutations formula to calculate how many ways there are to arrange objects when their order is important.
We’re gonna take a look at permutations and combinations. We’ll explain what they mean, and we’ll look at how we come up with the formulae, and we’ll also see how to use them. There are several different notations in common use around the world. I’m gonna use these ones here, but obviously you may have to use another format depending on where you live.
Right! Here’s a question. How many ways are there of arranging the letters 𝐴 and 𝐵? Well let’s think about this as we got two boxes to fill. The first, we pick a letter; the second, we pick a letter. Well there are obviously two choices for that first box. We could either pick an 𝐴 or we could pick a 𝐵, which just leaves us with one choice, the remaining letter, for the other box. So we had two choices for this one and one choice for this one, and each of those two could be combined with that one so basically we’re gonna multiply those two together. So two times one is a total of two different combinations. and obviously we could’ve had 𝐴 then 𝐵 or we could have 𝐵 then 𝐴. Okay that’s the trivial case. Let’s move on to a slightly more complicated case.
Okay it’s not much more complicated. How many ways are there of arranging the letters 𝐴, 𝐵, and 𝐶? So now we’ve got three boxes, and we’ve got three choices of letter for the first box: 𝐴 or 𝐵 or 𝐶. Now whatever letter we pick first, that leaves us with two choices for a second box. If we picked 𝐴, we can choose from 𝐵 and 𝐶. If we picked 𝐵 we can choose from 𝐴 and 𝐶. And if we picked 𝐶 first time round, we’ve then got 𝐴 and 𝐵 to choose from. So we’ve got one choice for the last box, and each of the three choices from the first box can be mixed with any of the two from the second, and they can be mixed with that last one as well which is a bit weird because there’s only one there. But what we can do is multiply those all together: three times two times one giving us a total of six combinations. So if we picked 𝐴 first, we can either have 𝐵 then 𝐶 or we could have 𝐶 then 𝐵. If we picked 𝐵 first, we could then have 𝐴 then 𝐶 or we could have 𝐶 then 𝐴. And if we pick 𝐶 first, the other two boxes contained-could contain 𝐴 then 𝐵 or 𝐵 then 𝐴. So there we have it, our six different permutations of arranging the letters 𝐴, 𝐵, and 𝐶 into three adjacent boxes.
Okay, just one more of these then. How many ways are there of arranging the four letters 𝐴, 𝐵, 𝐶, and 𝐷? So we’ve got four boxes now. So when we look at the first box, we’ve got four options to choose from: 𝐴, 𝐵, 𝐶, or 𝐷. And we pick one of those, and that only leaves us with three letters to choose from for the second box. We pick one of those; that leaves us with two for the third box which— It’s not really a choice but that only leaves one letter that we can put in the last box. So any of the first four choices can be combined with any of the three second choices which can be combined with any of the two sec- third choices which can be combined with the last choice as well. So basically four times three times two times one, twenty-four different permutations for those letters 𝐴, 𝐵, 𝐶, and 𝐷. So if I choose 𝐴 for my first box, I’ve then got 𝐵, 𝐶, and 𝐷 which I can organise for the other three boxes. Well we just looked at the three box problem, so there are six ways of organising those three letters. And if I chose 𝐵 as my first letter, that would leave me with 𝐴, 𝐶, and 𝐷 which again I’ve got six ways of distributing those across the other three boxes. And, again, another six each if I begin with a 𝐶 or a 𝐷.
Now as a slight aside, my personal favourite combination is this one here because it’s 𝐴𝐶𝐷𝐵, which is a chocolate digestive biscuit which is my favourite kind of biscuit. Anyway, moving on. So just summarising what we’ve done so far then: for two letters, the number of permutations is two times one; for three letters, it’s three times two times one; and for four letters, it’s four times three times two times one. So each time we add a new letter, we’re adding more permutations. So if we’ve got two letters, as we said it’s two times one. Well there are the two letters. If we add in another letter, we then have three choices for that first letter, and each of those three can be combined with the permutations for two letters that we had before. Likewise, we’ve now got three times two times one. So that was the number of permutations for three letters. If we add in a fourth letters, we have four choices for the first letter and then we’ve got that six choices- six different permutations for the other letters.
So if we had 𝑛 letters or let’s say 𝑛 objects, because there’s only twenty-six letters in the English alphabet so we’re talking this works for any number of objects so 𝑛 objects, to work out how many permutations for you can organise them, it’s that number 𝑛 times one less than that times one less than that times one less than that, and so on and so on and so on, times three times two times one all the way down to one. So it’s this formula here. And we’ve actually got a shorthand way of writing that. So to save writing out all those numbers every time, we’ve got this nice compact from here: 𝑛 factorial, 𝑛 with an exclamation mark after it, which means 𝑛 times the number that’s one smaller times the number that’s one smaller and so on and so on all the way down to one. In some parts of the world, you may see that written as this, and that means 𝑛 factorial but again that’s just different notation. The first one is the format that I’m gonna using in this video.
So in general then, for 𝑛 unique objects there are 𝑛 factorial ways of arranging them in different orders. And that’s easy to apply, so for example ah how many ways are there of arranging the letters 𝐴, 𝐵, 𝐶, 𝐷, 𝐸? Well there are five letters so there are five factorial ways of arranging them. That’s five times four times three times two times one. When we multiply all that together, we get one hundred and twenty different ways. Okay let’s refine the question slightly then and say how many ways are there to pick one letter from the four letters 𝐴, 𝐵, 𝐶, and 𝐷. Well we’ve got one box to fill. We’re picking one letter, and we’ve got four choices for that: 𝐴, 𝐵, 𝐶, or 𝐷. Okay then how many ways are there to pick two letters from the pile 𝐴, 𝐵, 𝐶, and 𝐷? Well we’ve got four to choose from for the first box and we’ve picked one then. So then you’ve got three to choose from for the second box. But any of those four can be combined with any of those three, so there are twelve different ways of doing this. If we picked 𝐴 first time, we could choose from 𝐵 or 𝐶 or 𝐷 for the second one. If we chose 𝐵 first, then we’ve got 𝐴, 𝐶, or 𝐷 to choose from. And similarly with 𝐶 and 𝐷, we have three different ways of combining those with the other letters. So there are twelve ways. So then how many ways to pick three letters from 𝐴, 𝐵, 𝐶, 𝐷? Well now we’ve got three boxes. We’ve got four choices for the first one, three choices for the second, and two choices for the third; they can all be combined together so we multiply them together. It’s twenty-four different choices. So if I picked 𝐴 first, I could then either pick 𝐵 or 𝐶 or 𝐷 second. And depending on which of those I pick, I’ve then got two more choices for that last option there. So for 𝐴 first, there are six different options; for 𝐵 first, likewise, there are six different options. For 𝐶 first, there are six more; and for 𝐷 first, there are six more, making a total of twenty-four different options or twenty-four permutations for this. Okay let’s just write those results down. So for one letter, we have four ways; two letters, it was four times three which gave us twelve permutations; three letters gave us four times three times two is twenty-four permutations; and four letters was four times three times two times one, which was twenty-four permutations. Now we didn’t actually go through that but that’s pretty obvious; once you’ve used the other three, you’ve only got one left so you only have one choice for that last place. So we’re just gonna set up a bit of notation then. So four 𝑃 one means how many permutations are there when we’re choosing one object out of four, four 𝑃 two means how many permutations are there when we’re choosing two objects out of four, four 𝑃 three means how many permutations for three objects out of four, and four 𝑃 four means how many permutations are there for choosing four objects out of four.
So we saw that there were four factorial ways of organising four letters, so four times three times two times one. So I’ve gotta think what can I do to that in order to get the answer for one letter when I’m choosing just one letter out of that four. Well, if I could cancel the three and the two and the one, that would just leave me with four. So when we’re looking at two letters, let’s think about that we’re trying to generate a calculation of four times three. Well if I do four times three times two times one, that’s the total number of combinations of four letters in- arranged in different ways. If I’m choosing two from that, the number of permutations, if I can cancel down the two and the one, so I cancel down the two and the one, that it would leave me with my calculation of four times three. And for three letters, we start off with the same number of combinations and we’re just gonna cancel down the one.
And I’ll come back and talk about the four-letter situation in just a moment. So let’s examine how we can generate these three calculations on the left. Well, the numerators are all four times three times two times one so they’re all four factorial, and the denominator in the first case is three times two times one. That’s three factorial. In the second case, it’s two times one which is two factorial. And in the third case, it’s one. Well, let’s just call that one factorial. And next so, we’ve got four on the top so that’s where this four comes from, and we’ve got three on the bottom. Well four take away one gives us three. For the second case, we have four on the top, that’s where that comes from. four take away two gives us two. And for the third one, we’ve got the four on the top and we’ve got one at the bottom, so four take away three leaves us with one.
So just trying to spot the pattern that’s emerging here, the number of objects that we had at the top here tells us what the number on the numerator is gonna be. And if we subtract the number that we’re choosing from that number, that number at the top there, that tells us what number is gonna go on the denominator. Now if we apply this pattern to the last case over here, we’ve got four factorial on the top and four minus four is zero factorial on the bottom. And there’s a bit of a problem, zero factorial. What on earth does that mean? Well remember, factorial is you take the number and then you keep subtracting away one and you’re multiplying. So four times three times two times one and so on. But if we start off at zero, what does that mean? Well there’s a bit of a trick here. We define that as being equal to one, and sorry about that! It might feel be a bit uncomfortable choosing to do that, but that’s the definition of zero factorial: it is one. So if we take that definition of zero factorial, we end up with four factorial over zero factorial, four factorial over one. So that’s still a number and it still gives us the right answer.
So 𝑛 𝑃 𝑟 is 𝑛 factorial over 𝑛 minus 𝑟 factorial, where 𝑛 𝑃 𝑟 is the number of permutations for organising 𝑟 objects from a set of 𝑛 objects. And remember this provisor here that we’ve defined zero factorial to be equal to one so that this whole thing works. So we’ve seen that there are twenty-four different permutations for picking three letters out of a bag of four letters, and these are the different combinations that we’ve written down here. So the thing you might notice is that some of them are actually equivalent. So for example, all these six different combinations are all just variations on 𝐴, 𝐵, and 𝐶, just in different orders. And these ones are the six variations of ways of organising 𝐵, 𝐶, and 𝐷. So those twenty-four different combinations in total actually boil down to six which are variations on 𝐴, 𝐵, and 𝐶; six which are variations on 𝐴, 𝐵, and 𝐷; six variations on 𝐴, 𝐶, and 𝐷; and six variations on 𝐵, 𝐶, and 𝐷.
So we’re gonna look at that situation in more detail in another video, where you are you’re not so interested in the order but you do want to know how many combinations of objects you get when pulling so many objects out of a bag.
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