![]() ![]() Now if we solve the above problem, we get total number of circular permutation of 3 persons taken all at a time = (3-1)! = 2. So for n elements, circular permutation = n! / n = (n-1)! ![]() Hence in general if we have n elements then total linear permutation of n elements taken all at a time is n! And we observe that n linear permutations correspond to 1 circular permutation. So it turns out that 3 linear permutations is actually 1 circular permutation. How? Well… If you move clockwise, start with A, round the table in the picture shown below you will always get A-B-C. If we arrange these 3 persons around a round table as show in the picture below, we notice that all the different arrangements are not actually different, rather they all are same. So we will have the following linear sitting arrangements Now for the sake of our convenience let us represent them as A, B and C. If we have 3 persons and if we want to arrange them in a linear fashion then the total number of permutation of 3 persons taken all at a time is 3P3 = 3! = 6. If we consider a round table and 3 persons then the number of different sitting arrangement that we can have around the round table is an example of circular permutation.Ĭircular permutation is a very interesting case. Permutation in a circle is called circular permutation. Therefore, the number of permutations in this case = 10x10x10x10x10x10 = 1000000 So we have 6 places and each of the places can be filled with any one of the 10 digits. Proceeding in a similar way C, D, E and F can be filled with any one of the 10 digits respectively. Similarly, we can fill the second place i.e., B with any one of the 10 digits as repetition is allowed so we can reuse the digit used in A. Note! While forming passwords even if we start with zero it will still count. We can fill the first place i.e., A with any one of the 10 digits. Let’s mark these places as A, B, C, D, E and F This means we have 6 places and 10 digits to fill those 6 places. ![]() So we have to find the number of permutations of 10 digits taken 6 at a time with repetition allowed. Say for instance, you have the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 and you are asked to find the total numbers of 6 digits passwords that can be formed using those 10 digits and repetition is allowed. This is a very interesting part of permutation. So the required number of permutation = 12! / (3! X 4! X 5!) = 27720 ways. Of which 3 are Harry Potter, 4 are The Lost Symbols by Dan Brown and 5 are The Secrete of The Unicorn by Herge. In how many ways can you arrange these books on a shelf?Ī. There are 3 copies of Harry Potter and the Philosopher's Stone, 4 copies of The Lost Symbol, 5 copies of The Secret of the Unicorn. Therefore, we can have 180 different arrangements. So total number of permutations in this case = 6! / (2! X 2!) = 180 In this case we have 6 letters of which 2 are O and 2 are G. In how many ways can the letters of the word “GOOGLE” be arranged?Ī. In this case the required number of permutations is written as, If we have n things of which x number of things are of same kind, y number of things are of same type and similarly z number of things are of the same type. Before diving into circular permutation let us discuss Permutation of n things not all different taken all together. ![]()
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