1. 10.1 - The Fundamental Counting Principal - Permutations - Factorials

2. The Fundamental Counting Principle If you have 2 events: 1 event can occur m ways and another event can occur n ways, then the number of ways that both can occur is m*n Event 1 = 4 types of meats Event 2 = 3 types of bread How many diff types of sandwiches can you make? 4*3 = 12

3. 3 or more events: 3 events can occur m, n, & p ways, then the number of ways all three can occur is m*n*p 4 meats 3 cheeses 3 breads How many different sandwiches can you make? 4*3*3 = 36 sandwiches

4. At a restaurant at Cedar Point, you have the choice of 8 different entrees, 2 different salads, 12 different drinks, & 6 different deserts. Question : How many different dinners can you choose (one choice of each)? 8*2*12*6= 1152 different dinners

5. Fundamental Counting Principle with Repetition Let’s say, Ohio Licenses plates have 3 #’s followed by 3 letters. How many different licenses plates are possible if digits and letters can be repeated? There are 10 choices for digits and 26 choices for letters. 10*10*10*26*26*26= 17,576,000 different plates

6. How many plates are possible if digits and numbers cannot be repeated? Same situation with license plates…but a number or letter cannot be repeated. There are still 10 choices for the 1 st digit, but only 9 choices for the 2 nd, and 8 for the 3 rd. For the letters, there are 26 for the first, but only 25 for the 2 nd and 24 for the 3 rd. 10*9*8*26*25*24 = 11,232,000 plates

7. Phone numbers How many different 7 digit phone numbers are possible if the 1 st digit cannot be a 0 or 1? 8*10*10*10*10*10*10= 8,000,000 different numbers

8. PASSWORDS Let’s pick a password…. a) 4 using digits only

9. PASSWORDS Let’s pick a password…. b) 4 using digits or letters (not case sensitive)

10. PASSWORDS Let’s pick a password…. c) 4 using digits or letters (upper/lower case sensitive)

11. Testing A multiple choice test has 10 questions with 4 answers each. How many ways can you complete the test? 4*4*4*4*4*4*4*4*4*4 = 4 10 = 1,048,576

12. Using Permutations An ordering of n objects is a permutation of the objects.

13. There are 6 permutations of the letters A, B, &C ABC ACB BAC BCA CAB CBA You can use the Fund. Counting Principle to determine the number of permutations of n objects. Like this ABC. There are 3 choices for 1 st # 2 choices for 2 nd # 1 choice for 3 rd. 3*2*1 = 6 ways to arrange the letters

14. In general, the # of permutations of n objects is: n! = n*(n-1)*(n-2)* …

15. 12 skiers… How many different ways can 12 skiers in the Olympic finals finish the competition? (if there are no ties) 12! = 12*11*10*9*8*7*6*5*4*3*2*1 479,001,600 different ways

16. Factorial with a calculator: Hit math then over, over, over. Option 4

17. Back to the finals in the Olympic skiing competition. How many different ways can 3 of the skiers finish 1 st, 2 nd, & 3 rd (gold, silver, bronze) Any of the 12 skiers can finish 1 st, the any of the remaining 11 can finish 2 nd, and any of the remaining 10 can finish 3 rd. So the number of ways the skiers can win the medals is 12*11*10 = 1320

18. Permutation of n objects taken r at a time n P r =

19. Back to the last problem with the skiers It can be set up as the number of permutations of 12 objects taken 3 at a time. 12 P 3 = 12! = 12! = (12-3)!9! 12*11*10*9*8*7*6*5*4*3*2*1 = 9*8*7*6*5*4*3*2*1 12*11*10 = 1320

20. 10 colleges, you want to visit all or some. How many ways can you visit 6 of them: Permutation of 10 objects taken 6 at a time: 10 P 6 = 10!/(10-6)! = 10!/4! = 3,628,800/24 = 151,200

21. How many ways can you visit all 10 of them: 10 P 10 = 10!/(10-10)! = 10!/0!= 10! = ( 0! By definition = 1) 3,628,800

22. How many different ways can we arrange the letters A, B, &C ABC ACB BAC BCA CAB CBA 3!

23. If some of the objects are repeated, then some of the permutations are not distinguishable.  There are 6 ways to order the letters M,O,M  MOM, OMM, MMO  Only 3 are Distinguishable.

24. Find the number of Distinguishable permutations of the letters: OHIO : 4 letters with O repeated 2 times 4! = 24 = 12 2! 2 MISSISSIPPI : 11 letters with I repeated 4 times, S repeated 4 times, P repeated 2 times 11! = 39,916,800 = 34,650 4!*4!*2! 24*24*2

25.  SUMMER  360  WATERFALL  90,720

27. Permutations with Repetition The number of DISTINGUISHABLE permutations of n objects where one object is repeated q 1 times, another is repeated q 2 times, and so on : n! q 1 ! * q 2 ! * … * q k !

28. A dog has 8 puppies, 3 male and 5 female. How many birth orders are possible by gender. 8!/(3!*5!) = 56

29. Assignment