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# 12 1 The Fundamental Counting Principal & Permutations Revised

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### 12 1 The Fundamental Counting Principal & Permutations Revised

1. 1. 12.1 The Fundamental Counting Principal & Permutations
2. 2. The Fundamental Counting Principal <ul><li>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 </li></ul><ul><li>Event 1 = 4 types of meats </li></ul><ul><li>Event 2 = 3 types of bread </li></ul><ul><li>How many diff types of sandwiches can you make? </li></ul><ul><li>4*3 = 12 </li></ul>
3. 3. 3 or more events: <ul><li>3 events can occur m, n, & p ways, then the number of ways all three can occur is m*n*p </li></ul><ul><li>4 meats </li></ul><ul><li>3 cheeses </li></ul><ul><li>3 breads </li></ul><ul><li>How many different sandwiches can you make? </li></ul><ul><li>4*3*3 = 36 sandwiches </li></ul>
4. 4. <ul><li>At a restaurant at Cedar Point, you have the choice of 8 different entrees, 2 different salads, 12 different drinks, & 6 different deserts. </li></ul><ul><li>How many different dinners (one choice of each) can you choose? </li></ul><ul><li>8*2*12*6= </li></ul><ul><li>1152 different dinners </li></ul>
5. 5. Fund. Counting Principal with repetition <ul><li>Ohio Licenses plates have 3 #’s followed by 3 letters. </li></ul><ul><li>1. How many different licenses plates are possible if digits and letters can be repeated? </li></ul><ul><li>There are 10 choices for digits and 26 choices for letters. </li></ul><ul><li>10*10*10*26*26*26= </li></ul><ul><li>17,576,000 different plates </li></ul>
6. 6. How many plates are possible if digits and numbers cannot be repeated? <ul><li>There are still 10 choices for the 1 st digit but only 9 choices for the 2 nd , and 8 for the 3 rd . </li></ul><ul><li>For the letters, there are 26 for the first, but only 25 for the 2 nd and 24 for the 3 rd . </li></ul><ul><li>10*9*8*26*25*24= </li></ul><ul><li>11,232,000 plates </li></ul>
7. 7. Phone numbers <ul><li>How many different 7 digit phone numbers are possible if the 1 st digit cannot be a 0 or 1? </li></ul><ul><li>8*10*10*10*10*10*10= </li></ul><ul><li>8,000,000 different numbers </li></ul>
8. 8. Testing <ul><li>A multiple choice test has 10 questions with 4 answers each. How many ways can you complete the test? </li></ul><ul><li>4*4*4*4*4*4*4*4*4*4 = 4 10 = </li></ul><ul><li>1,048,576 </li></ul>
9. 9. Using Permutations <ul><li>An ordering of n objects is a permutation of the objects . </li></ul>
10. 10. There are 6 permutations of the letters A, B, &C <ul><li>ABC </li></ul><ul><li>ACB </li></ul><ul><li>BAC </li></ul><ul><li>BCA </li></ul><ul><li>CAB </li></ul><ul><li>CBA </li></ul>You can use the Fund. Counting Principal 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
11. 11. In general, the # of permutations of n objects is: <ul><li>n! = n*(n-1)*(n-2)* … </li></ul>
12. 12. 12 skiers… <ul><li>How many different ways can 12 skiers in the Olympic finals finish the competition? (if there are no ties) </li></ul><ul><li>12! = 12*11*10*9*8*7*6*5*4*3*2*1 = </li></ul><ul><li>479,001,600 different ways </li></ul>
13. 13. Factorial with a calculator: <ul><li>Hit math then over, over, over. </li></ul><ul><li>Option 4 </li></ul>
14. 14. Back to the finals in the Olympic skiing competition. <ul><li>How many different ways can 3 of the skiers finish 1 st , 2 nd , & 3 rd (gold, silver, bronze) </li></ul><ul><li>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 . </li></ul><ul><li>So the number of ways the skiers can win the medals is </li></ul><ul><li>12*11*10 = 1320 </li></ul>
15. 15. Permutation of n objects taken r at a time <ul><li>n P r = </li></ul>
16. 16. Back to the last problem with the skiers <ul><li>It can be set up as the number of permutations of 12 objects taken 3 at a time. </li></ul><ul><li>12 P 3 = 12! = 12! = (12-3)! 9! </li></ul><ul><li>12*11*10*9*8*7*6*5*4*3*2*1 = 9*8*7*6*5*4*3*2*1 </li></ul><ul><li>12*11*10 = 1320 </li></ul>
17. 17. 10 colleges, you want to visit all or some. <ul><li>How many ways can you visit </li></ul><ul><li>6 of them: </li></ul><ul><li>Permutation of 10 objects taken 6 at a time: </li></ul><ul><li>10 P 6 = 10!/(10-6)! = 10!/4! = </li></ul><ul><li>3,628,800/24 = 151,200 </li></ul>
18. 18. How many ways can you visit all 10 of them: <ul><li>10 P 10 = </li></ul><ul><li>10!/(10-10)! = </li></ul><ul><li>10!/0!= </li></ul><ul><li>10! = ( 0! By definition = 1) </li></ul><ul><li>3,628,800 </li></ul>