![Page 1: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/1.jpg)
12.1The Fundamental Counting Principal & Permutations
![Page 2: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/2.jpg)
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 meatsEvent 2 = 3 types of bread
How many diff types of sandwiches can you make?
4*3 = 12
![Page 3: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/3.jpg)
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 meats3 cheeses3 breadsHow many different sandwiches can you make?
4*3*3 = 36 sandwiches
![Page 4: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/4.jpg)
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
![Page 5: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/5.jpg)
Fund. Counting Principal 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
![Page 6: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/6.jpg)
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
![Page 7: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/7.jpg)
Phone numbersHow 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
![Page 8: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/8.jpg)
TestingA 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
![Page 9: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/9.jpg)
Using Permutations
An ordering of n objects is a permutation of the objects.
![Page 10: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/10.jpg)
There are 6 permutations of the letters A, B, &C
ABCACBBACBCACABCBA
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
![Page 11: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/11.jpg)
In general, the # of permutations of n objects
is:
n! = n*(n-1)*(n-2)* …
![Page 12: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/12.jpg)
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
![Page 13: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/13.jpg)
Factorial with a calculator:
•Hit math then over, over, over.•Option 4
![Page 14: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/14.jpg)
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
![Page 15: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/15.jpg)
Permutation of n objects taken r at a time
nPr =
!
!
rn
n
![Page 16: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/16.jpg)
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
![Page 17: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/17.jpg)
10 colleges, you want to visit all or some.How many ways can you visit6 of them:
Permutation of 10 objects taken 6 at a time:
10P6 = 10!/(10-6)! = 10!/4! =
3,628,800/24 = 151,200
![Page 18: 12 1 The Fundamental Counting Principal & Permutations Revised](https://reader033.vdocument.in/reader033/viewer/2022061114/545cea3db0af9fa42c8b4a01/html5/thumbnails/18.jpg)
How many ways can you visitall 10 of them:
10P10 =
10!/(10-10)! = 10!/0!=10! = ( 0! By definition = 1)
3,628,800