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Unformatted text preview: Lecture 20 Andrei Antonenko March 26, 2003 1 Permutations Let we have n numbers from 1 to n : 1 , 2 ,...,n . If we change their order we get their per mutation . We will write these numbers in brackets, for example, (34152) is a permutation of numbers from 1 to 5. Example 1.1. There are 6 different permutations of numbers 1 , 2 and 3 : (123) (132) (213) (231) (312) (321) Lets compute the total number of permutations of numbers from 1 to n . On the first place in the permutation we can put any of n elements. On the second place we can put any element which is not equal to the element on the first place so we have ( n 1) possibilities; than to the third place we can put any element which is not on the first and the second place, so totally we have ( n 2) possibilities, etc. Finally, we will have only one possibility to put an element on the last nth place. So, total number of permutations of n elements is n ( n 1)( n 2) 2 1 = n ! (By definition, n ! is a product of all numbers from 1 to n . It is called nfactorial ). So, we see that the number of permutations increases very fast. We have just 3! = 6 different permutations of numbers from 1 to 3, 4! = 24 different permutations of numbers from 1 to 4,permutations of numbers from 1 to 3, 4!...
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.
 Spring '03
 Andant

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