# Video: Finding Probability Using Permutations

Tim Burnham

Eugene received 10 letters in the mail. If he can read them in any order, determine the probability that he reads the letter from his parents and then that from his grandmother before the others.

05:54

### Video Transcript

Matthew received 10 letters in the mail. If he can read them in any order, determine the probability that he reads the letter from his parents and then that from his grandmother before the others.

So there is ten letters sitting on the doormat. And let’s state a big assumption, that he’s gonna pick each letter to read at random. Let’s face it. If he recognised the handwriting from his parents and his grandmother, it will be perfectly possible for him to organise, to definitely read them in that particular order. So the probability will be one. Of course, that’s a perfectly acceptable answer. But the sad truth is, in your Masters, you probably won’t get any marks for that kind of answer.

So this is the assumption that we’re gonna be working with. Now we’re gonna answer this question using permutations, and then we’re gonna check it using a slightly different method. Now recall, the probability that an event occurs is equal to the number of ways that that event can occur divided by the total number of ways that things can happen. In other words, what proportion of cases of possible things result in our event occurring. And in this particular case, the event occurring means that he reads his parents letter first, his grandmother letter second, and then all of the other letters in any order after that. So when objects are arranged and the order in which they are arranged is important, we call it a permutation. And there are ten factorial permutations for how Matthew could read the letters.

Look, when Matthew chooses his first letter, he’s got ten to choose from. When he chooses his second letter, he’s already read one. So there’re only nine to choose from. For the third, there’re only eight left to choose from, and so on. Now any of those first ten choices can be combined with any of those second nine choices. And all of those choices can be combined with any of the next eight choices.

So if you want to work out the total number of different orders in which he could read the letters, that’s ten times nine times eight times seven times six times five times four times three times two times one. And we’ve got a quicker way of saying that, ten factorial. And when you work that out, you get three million six hundred and twenty-eight thousand eight hundred different orders in which he could read the letters. So the total number of different ways of these things can happen in this case is three million six hundred and twenty-eight thousand eight hundred.

But in how many of those different permutations did he read his parents letter first and his grandparents letter second? Well, let’s take his parents letter out of the poll and put that in the first slot and his grandmother’s letter out of the poll and put that in the second slot. So how many ways can the rest of his reading pan out? Well for the third slot, he’s now got a choice of eight different letters. For the fourth slot, he’s only got seven. The fifth slot, six and so on and so on down to just one remaining for his last letter. Given that we fix this so that he definitely read his parents letter first and his grandmother’s letters second, that means we’ve got eight times seven times six times five times four times three times two times one ways of reading the other letters, or eight factorial ways. Now eight factorial is equal to forty thousand three hundred and twenty. So there are forty thousand three hundred and twenty different permutations for how he could read the other eight letters.

So these are the number of ways in which we force him to read the parents letter first, the grandmother’s letter second. And these are the total number of ways that things can happen if we don’t put that restriction on him. So remember, the probability that he reads his parents letter first then his grandmother’s letter and then all of the rest in any order is equal to the number of ways that this can happen divided by the total number of permutations we’ve got, which we said was eight factorial over ten factorial. So that’s forty thousand three hundred and twenty over three million six hundred and twenty-eight thousand eight hundred, which is one over ninety. So the probability he reads parents letter first then his grandmother’s and then all the others is one ninetieth.

Now just before we go on and do a check, let’s find an alternative way of getting from here to here. But remember, eight factorial is eight times seven times six times five times four times three times two times one and ten factorial is ten times nine times eight times seven times six times five times four times three times two times one. So you can see that the eight on the top cancels with the eight on the bottom, the seven on the top with the seven on the bottom and so on all the way down to one, until we’re left with just one on the top and ten times nine on the bottom. And ten times nine is ninety.

Right. Now let’s try and think of an alternative approach to solving this problem. If we think about it, the only critical things there are that he picks the parents letter first and then the grandmother’s letter second. And then we don’t really care about the r- the order of the rest of the letters. So we could say that’s the same as the probability that he gets the parents first and the grandmother’s second. So when he is randomly picking the first letter, one of those ten letters is his parents letter. So he’s got a one in ten chance of picking that. And for the second one, he’s got nine letters left. One of which is from his grandmother. So there’s a one in nine chance of picking his grandmother’s letter. Now because we want this and this to be true, we have to multiply those probabilities together. And when you multiply fractions together, you multiply the numerators, multiply the denominators. So that’s one over ninety.

So using a different method, we’ve got the same answer. So we’re pretty confident it’s correct.