By René Schoof
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Extra resources for Algebra 2. The symmetric groups Sn
3628800, 11 ! = 39916800. 10. Enumerating r-permutations without repetitions P(n, r) = n(n – 1) ...... (n – r + 1) = n! (n − r ) ! Proof. Since there are n distinct objects, the first position of an r-permutation may be filled in n ways. This done, the second position can be filled in n – 1 ways since no repetitions are allowed and there are n – 1 objects left to choose from. The third can be filled in n – 2 ways and so on until the rth position is filled in n – r + 1 ways . (See figure below).
The number of sequence is the sum of coefficients of all the terms in this generating function. This is equal to 64. 65. Find the generating function, also called enumerator, for permutations of n objects with the following specified conditions : (a) each object occurs at the most twice (b) each object occurs at least twice (c) each object occurs at least once and at the most k times. Solution. (a) Each object occurs at the most twice implies that an object may occur 0, 1 or 2 times. The exponential generating function for an object under this condition is given as 1+x+ x2 .
Therefore, the number of ways of scheduling 10 papers so that the best and the worst never come together = 10 ! – 725760 = 3628800 – 725760 = 2903040. 59. In how many different ways can 5 men and 5 women sit around a table, if (i) there is no restriction (ii) no two women sit together ? Solution. The problem is related to circular permutation of 10 objects (5 men and 5 women). If there is no restriction then the number of permutations is (10 – 1) ! = 9 ! = 362880. Notice here the difference in arrangement between clockwise and anticlockwise.