Let's say that we only have three chairs. We have chair one, we have chair two, and we have chair three. How many ways can you have five people, where only three of them are going to sit down in these three chairs, and we care which chair they sit in? I encourage you to pause the video and think about it. I am assuming you have had your go at it. Let's use the same logic. If we seat them in order, and we might as well, how many different people, if we haven't sat anyone yet, how many different people could sit in seat one?
Well, we could have, if no one sat down, we had five different people, five different people could sit in seat one. For each of these scenarios where one person has already sat in seat one, how many people could sit in seat two? In each of these scenarios, if one person has sat down, there's four people left who haven't been seated, so four people could sit in seat two.
So we have five times four scenarios where we've seated seats one and seat two. For each of those 20 scenarios, how many people could sit in seat three? Well, we haven't sat, we haven't seaten or sat three of the people yet, so for each of these 20, we could put three different people in seat three, so that gives us five times four times three scenarios.
So this is equal to five times four times three scenarios, which is equal to, this is equal to So there's 60 permutations of sitting five people in three chairs. Now this, and my brain, whenever I start to think in terms of permutations, I actually think in these ways. I just literally draw it out because I don't like formulas. I like to actually conceptualize and visualize what I'm doing. But you might say, hey, when we just did five different people in five different chairs, and we cared which seat they sit in, we had this five factorial.
Factorial is kind of neat little operation there. How can I relate factorial to what we did just now? It looks like we kind of did factorial, but then we stopped. How many different ways are there of selecting the three balls? Since the order is important, it is the permutation formula which we use. In the National Lottery, 6 numbers are chosen from You win if the 6 balls you pick match the six balls selected by the machine.
What is the probability of winning the National Lottery? Skip to main content. There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:. Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.
So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on.
We get. The problem is easily solved by the multiplication axiom, and answers are as follows:. We often encounter situations where we have a set of n objects and we are selecting r objects to form permutations.
We refer to this as permutations of n objects taken r at a time , and we write it as nPr. Before we give a formula for nPr, we'd like to introduce a symbol that we will use a great deal in this as well as in the next chapter. So far we have used up 2 letters, therefore, five remain.
So for the next position there are five choices, for the position after that there are four choices, and so on. We get:. Factorial : n! The reader should become familiar with both formulas and should feel comfortable in applying either. Next we consider some more permutation problems to get further insight into these concepts. Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5 P 2 permutations of history books that can be placed in the last two slots.
Hence the multiplication axiom applies, and we have the answer 4 P 3 5 P 2. We summarize. The first problem comes under the category of Circular Permutations , and the second under Permutations with Similar Elements. Suppose we have three people named A , B , and C. We have already determined that they can be seated in a straight line in 3! Our next problem is to see how many ways these people can be seated in a circle.
We draw a diagram:. It happens that there are only two ways we can seat three people in a circle. This kind of permutation is called a circular permutation. In such cases, no matter where the first person sits, the permutation is not affected. Each person can shift as many places as they like, and the permutation will not be changed. Imagine the people on a merry-go-round; the rotation of the permutation does not generate a new permutation.
So in circular permutations, the first person is considered a place holder, and where he sits does not matter. We again emphasize that the first person can sit anywhere without affecting the permutation. So there is only one choice for the first spot.
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