The Fundamental Counting Principal Permutations The Fundamental Counting
- Slides: 24
The Fundamental Counting Principal & Permutations
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 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
• 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
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
How many plates are possible if digits and numbers 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
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= • 8, 000 different numbers
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 = 410 = • 1, 048, 576
Using Permutations • An ordering of n objects is a permutation of the objects.
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 1 st # 2 choices for 2 nd # 1 choice for 3 rd. 3*2*1 = 6 ways to arrange the letters
In general, the # of permutations of n objects is: • n! = n*(n-1)*(n-2)* …
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
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
Permutation of n objects taken r at a time P = • n r
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 -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
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
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
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 • Only 3 are distinguishable. 3!/2! = 6/2 = 3
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! * … * qk!
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
Find the number of distinguishable permutations of the letters: • SUMMER : • 360 • WATERFALL : • 90, 720
A dog has 8 puppies, 3 male and 5 female. How many birth orders are possible • 8!/(3!*5!) = • 56
Combination of n objects taken r at a time C = • n r Combination is used when order of events IS NOT important!
EXAMPLE • Your favorite pizza place has a choice of 8 different toppings. You’ve pooled enough cash from your friends for a pizza with 3 toppings. • How many different pizzas can you order?
- Permutation counting principle
- Fundamental counting principle
- Counting rules in probability
- Symbolic probability rules
- Fundamental counting principle example
- Permutations and combinations
- Fundamental counting principle and factorial notation
- Tree diagram outcomes
- Kotak 1 berisi 2 bola merah dan 3 bola putih
- Permutations
- Fundamental counting principle
- What is the fundamental principle of counting
- Counting techniques tree diagram
- Fundamental principles of counting
- Table
- Fundamental counting principle definition
- Generalized permutations and combinations
- Combination examples
- Multiset combinations
- Permutation formula
- Circular permutation with restriction
- Permutasi string
- Permutation examples
- Generalized permutations and combinations
- Permutations