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9.6 The Fundamental Counting Principal & Permutations

Explore the Fundamental Counting Principle with examples in combinations and permutations, including factorial calculations and applying permutations with repetition to various scenarios to comprehend ordering and arrangements more effectively.

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9.6 The Fundamental Counting Principal & Permutations

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  1. 9.6The Fundamental Counting Principal & Permutations

  2. The Fundamental Counting Principal • 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 desserts. • How many different dinners (one choice of each) can you choose? • 8*2*12*6= • 1152 different dinners

  5. Fund. Counting Principle with repetition • Ohio License plates have 3 #’s followed by 3 letters. • 1. 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 letters cannot be repeated? • There are still 10 choices for the 1st digit but only 9 choices for the 2nd, and 8 for the 3rd. • For the letters, there are 26 for the first, but only 25 for the 2nd and 24 for the 3rd. • 10*9*8*26*25*24= • 11,232,000 plates

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

  8. 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 = 410 = • 1,048,576

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

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

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

  12. 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

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

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

  15. Permutation of n objects taken r at a time • nPr =

  16. 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. • 12P3 = 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

  17. 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: • 10P6 = 10!/(10-6)! = 10!/4! = • 3,628,800/24 = 151,200

  18. How many ways can you visitall 10 of them: • 10P10 = • 10!/(10-10)! = • 10!/0!= • 10! = ( 0! By definition = 1) • 3,628,800

  19. So far in our problems, we have used distinct objects. • 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 • MOM, OMM, MMO • Only 3 are distinguishable. 3!/2! = 6/2 = 3

  20. Permutations with Repetition • The number of DISTINGUISHABLE permutations of n objects where one object is repeated q1 times, another is repeated q2 times, and so on : • n! q1! * q2! * … * qk!

  21. Find the number of distinguishable permutations of the letters: • OHIO : 4 letters with 0 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

  22. Find the number of distinguishable permutations of the letters: • SUMMER : • 360 • WATERFALL : • 90,720

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

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