1. Permutations

2. Basics 5! Does not mean FIVE! 5! = 5 4 3 2 1 7! = 7 6 5 4 3 2 1 5! is read five factorial. 5! = 120 7! = 5040

3. When a group of objects or things are arranged in a certain order, the arrangement is called a permutation. The arrangement of objects in a line is called a linear permutation.

4. Example 1. A theater owner has 11 films to show on 8 different screens. How many different arrange- ments are there for the 8 screens to show the 11 movies.

5. Example 1. 8 screens and 11 movies. We must make 8 choices. 1234 5678 111098 7654 Choice Choice # #

6. Example 1. 1234 5678 111098 7654 Choice Choice # # Therefore by the FCP there are 11 10 9 8 7 6 5 4

7. Example 1. Therefore by the FCP there are 11 10 9 8 7 6 5 4 arrangements. The number of way to arrange 11 objects taken 8 at a time is written as P(11,8).

8. P(n,r) is read as the permutation of n objects taken r at a time. P(n,r) = Definition of P(n,r) n! n!(n-r)!

9. Example 2. P(7,5) = Five teens find seven empty seats at a theater. How many different seating arrangements are there? 7! 7!(7-5)! 7!2!=

10. Example 2. P(7,5) = 7! 7!(7-5)! 7!2!= 7 6 5 4 3 2 1 2 1 2 1= = 7 6 5 4 3 = 2520

11. Example 3. A music store manager want to arrange 5 rock CD’s 4 rap CD’s and 4 jazz CD’s on a shelf. How many ways can they be arranged if they are ordered according to type?

12. Example 3. The 5 rock CD’s arrange into P(5,5) or 5! ways. The 4 rap CD’s arrange into P(4,4) or 4! ways. The 4 jazz CD’s arrange into P(4,4) or 4! ways.

13. Example 3. There are 3 types of CD’s that arrange into P(3,3) or 3! ways. The total ways the CD’s can be arranged is the product. P(5,5)P(4,4)P(4,4)P(3,3) = 5!4!4!3! = 414,720

14. Permutations with repetitions. The number of permutations of n objects of which p are alike and q are alike is n! n! p!q! p!q! The rule can be extended to any number of objects that repeat.

15. Example 4. How many ways can the letters of the word perpendicular be arranged? There are 2 p’s, 2 e’s, and 2 r’s. Therefore 13! 13!2!2!2!arrangements

16. Example 4. How many ways can the letters of the word perpendicular be arranged? 13! 13!2!2!2! = 778,377,600

17. Circular Permutations. If n distinct objects are arranged in a circle there are or (n-1)! permutations of the objects around the circle. n!n

18. Example 5. A disc jockey is loading a circular tray with six compact discs. How many different ways can these be arranged? (n-1)! = (6-1)! = 5! = 120

19. Circular Permutations (cont). If n distinct objects are arranged in a circle and there is a fixed point on the circle then there are n! permutations of the objects around the circle.

20. Example 6. If five different homes are being built around a cul-de-sac. How many different arrangements of the homes are possible? The cul-de-sac is circular but the entrance road is a fixed point

21. Example 6. If five different homes are being built around a cul-de-sac. The cul-de-sac is circular but the entrance road is a fixed point There are n! = 5! =120 arrangements possible.

22. We now have several formulas to remember. Linear permutations, circular permutations, circular permutations with a fixed point, and permutations with repetitions.

23. One last thing we need to know. Reflection happens when a circular permutation can be flipped over or when a linear permutation is viewed from opposite sides. This causes the number of permutations to be half as many.