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The Fundamental Counting Principle & Permutations

The Fundamental Counting Principle & Permutations. The Fundamental Counting Principle & Permutations Essential Question. How is the counting principle applied to determine outcomes?. The Fundamental Counting Principle.

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The Fundamental Counting Principle & Permutations

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  1. The Fundamental Counting Principle & Permutations

  2. The Fundamental Counting Principle & Permutations Essential Question How is the counting principle applied to determine outcomes?

  3. 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

  4. 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

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

  6. Fundamental Counting Principle with repetition Ohio Licenses 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

  7. How many plates are possible if digits and numbers 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

  8. 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

  9. 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

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

  11. There are 6 permutations of the letters A, B, &C • ABC • ACB • BAC • BCA • CAB • CBA You can use the Fundamental Counting Principle 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

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

  13. 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

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

  15. 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

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

  17. 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

  18. 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

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

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