![]() ![]() Together we will work through numerous examples, increasing in difficulty as we go, so we make sense of arranging elements in order. In other words, the number of 4-letter arrangements without repetition using the letters from the word GRAM is 4*3*2*1 = 24. 'circle, wheel, any circular body,' also 'circular motion, cycle of events,' Example 1. The etymology of the word 'cyclic' or 'cycle' comes from the Greek 'kyklos' which means. This would mean that if we wanted to construct a 4-letter word only using the letters from the word GRAM, we have 4 choices for the first letter, but once it’s used, we can’t choose it again, so that leaves 3 options for the second letter and 2 choices for the third and 1 choice for the fourth. With this terminology, a circular permutation is just exactly a permutation consisting of a single cycle that permutes all of the objects. Now, let’s assume repetition is not allowed. This means we are dealing with an r-permutation with repetition, so we would say there are 4*4*4*4=256 possible 4-letter arrangements. Well, there are 4 letters in the word GRAM so that means, if we wanted to, we could say GGGG or RRRR or AMAM. Okay, so now let’s suppose we are given the word GRAM, and we are asked to find the number of arrangements that can be made ( they don’t have to spell a word) from the letters in the word GRAM if repetition is allowed. Let’s write our problem using mathematical notation. Permutation Notationĭo you realize that you just used permutations? So, the total number of picture arrangements given 5 pictures for 3 spots is 5*4*3 = 60. Let’s let D be chosen.Īnd our third and final spot on the wall has 3 choices (B, D, E) as pictures A and D have now been hung up. The second spot on the wall has 4 choices (B, C, D, E), as picture A has already been hung. Let’s say A is the picture that you choose. So, for the first spot on the wall, there are 5 possible pictures to choose from (A, B, C, D, E). This implies that repetition is not allowed, and placement (order) matters. But what can’t happen is something like this: AAA or BBB or ABA, etc., because once we hang picture A on the wall, we can’t hang it somewhere else as it’s already been used. This means each of the five photos can be in first, second, or third place on the wall. This means we can arrange the three images: ABC or ABD or ABE or BCA or CDE, and so on. Let’s say the 5 pictures are labeled A, B, C, D, and E for ease of purpose. How many different arrangements can be made? Let’s suppose you have 5 pictures, and you want to arrange 3 of them on your wall. Let’s look at a few examples so we can get a better understanding of how permutations work. Hence the multiplication axiom applies, and we have the answer (4P3) (5P2).So, what this indicates is that when we aren’t allowed repetition, we must reduce the number of ways (choices) to arrange the objects. For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. So the answer can be written as (4P3) (5P2) = 480.Ĭlearly, this makes sense. Therefore, the number of permutations are \(4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 = 480\).Īlternately, we can see that \(4 \cdot 3 \cdot 2\) is really same as 4P3, and \(5 \cdot 4\) is 5P2. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The fourth slot requires a history book, and has five choices. ![]() Since the math books go in the first three slots, there are 4 choices for the first slot,ģ choices for the second and 2 choices for the third. We first do the problem using the multiplication axiom. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books? You have 4 math books and 5 history books to put on a shelf that has 5 slots. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements The multiplication axiom tells us that three people can be seated in 3! ways. Let us now do the problem using the multiplication axiom.Īfter we tie two of the people together and treat them as one person, we can say we have only three people. ![]() So altogether there are 12 different permutations.
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