1 If a task is made up of multiple operations (activities or stages that are independent of each other) the total number of possibilities for the multi-step task is given by m x n x p x... arrangements of repeats Fundamental Counting Principle A permutation determines the number of ways to list or arrange items. A permutation is a linear arrangement of a set of objects for which the order of the objects is important. Items may be identical or may repeat. A combination determines the number of ways to group items. Items must be unique and may not be repeated. Many permutations question may be computed by using FCP. 11.1B Permutations

Math Select the appropriate strategy for calculating each situation choosing 3 friends to go out to lunch determine arrangements of letters arranging vases on a shelf picking your favourite books to make a Top 10 List picking and ranking your favourite books on a Top 10 List

In how many ways can the letters of the word HARMONY be arranged? 7! = 5040 In how many ways can the letters of the word HARMONY be arranged if the arrangement must start with H? 1 x 6! = 720 In how many ways can the letters of the word HARMONY be arranged if the letters H and A must be together (adjacent)? = ! x 6! Math The number of permutations of n different objects taken all at a time is n!

How many three-letter arrangements can be formed from the letters of the word DINOSAUR? ____ x ____ x ____ 1 st 2 nd 3 rd 8 76 There would be 336 ways. 336 represents the number of permutations of eight objects taken three at a time. 8P38P3 is read as “eight permute three”. In general, if we have n objects but only want to select r objects at a time, the number of different linear arrangements is: Finding the Number of Permutations Write 8 x 7 x 6 in factorial notation. What restrictions are on n and r? 4

Finding the Number of Permutations The number of permutations of n objects taken n at a time is: n P n = n! 3P33P3 From these two results, we see that 0! = 1. To have meaning when r = n, we define 0! = 1. = 3! Math 30-15

1. Using the letters of the word PRODUCT, how many four-letter arrangements can be made? 7P47P4 = 840 There would be 840 arrangements. Finding the Number of Permutations 2. Determine the number of different arrangements of four or more letters that can be formed with the letters of the word LOGARITHM if each letter is not used more than once. 9P49P4 + 9 P P P P P 9 = There would be arrangements. Math True or False

4. How many six-letter words can be formed from the letters of TRAVEL? (Note that letters cannot be repeated) a) If any of the six letters can be used. 6 P 6 = 6! = 720 b) If the first letter must be “L”: 1 x 5 P 5 = 1 x 5! = 120 c) If the second and fourth letters are vowels: ___ ___ ___ V V 21 2 x 1 x 4! = 48 d) If the “A” and the “V” must be adjacent: (Treat the AV as one group - this grouping can be arranged 2! ways.) 5! x 2! = 240 Finding the Number of Permutations Math 30-17

Finding the Number of Permutations….. with Repeats How many arrangements are there of four letters from the word PREACHING? 9P49P4 How many distinct arrangements of BRAINS are there keeping the vowels together? 5! x 2! = 240 Math = 3024

A bookshelf contains five different algebra books and seven different physics books. How many different ways can these books be arranged if the algebra books are to be kept together? Total number of arrangements = 8! x 5! = Finding the Number of Permutations (Grouping) A student has 4 different biology books, 5 different chemistry books, and 6 different math books. In how many ways can the books be arranged so that the biology books stand together, the chemistry books stand together and the math books stand together. 4! x 5! x 6! x 3!= There are ways of arranging the books. Math 30-19

10 Determine the number of different arrangements using all the letters of the word ACCESSES that begin with at least two S’s

Two objects are selected from a group and arranged in order. If there are 90 possible arrangements, how many objects are there? n P 2 = 90 n(n - 1) = 90 n 2 - n = 90 n 2 - n - 90 = 0 (n - 10)(n + 9) = 0 n - 10 = 0 n = 10 OR n + 9 = 0 n = -9 n  N Therefore n = 10. Permutations 11

Solving Equations Involving Permutations. Solve n P 2 = 30 algebraically. n(n - 1) = 30 n 2 - n = 30 n 2 - n - 30 = 0 (n - 6)(n + 5) = 0 n = 6 or n = -5 Therefore, n = 6. 12

Solving Equations Involving Permutations. Solve n P 4 = 8( n-1 P 3 ) algebraically. n = 8 Math

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