Download presentation

Presentation is loading. Please wait.

Published byCarlton Covel Modified over 3 years ago

1
**The Fundamental Counting Principle & Permutations**

2
**Why do you use a fundamental counting principal?**

What operation do you use for fundamental counting principals? What is a permutation? What is the formula for nPr? What is the formula for permutations with repetition?

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 different 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
**Fund. 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
**Introduction to Permutations**

12
**There are 6 permutations of the letters A, B, &C**

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

13
**In general, the # of permutations of n objects is:**

n! = n*(n-1)*(n-2)* …

14
**12 skiers… 12! = 12*11*10*9*8*7*6*5*4*3*2*1 =**

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

15
**Factorial with a calculator:**

Hit math then over, over, over. Option 4

17
**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

18
**Permutation of n objects taken r at a time**

nPr =

19
**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 = *8*7*6*5*4*3*2*1 12*11*10 = 1320

20
**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

21
**How many ways can you visit all 10 of them:**

10P10 = 10!/(10-10)! = 10!/0!= 10! = ( 0! By definition = 1) 3,628,800

22
**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

23
**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 :

24
**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, = 34,650 4!*4!*2! 24*24*2

25
**Find the number of distinguishable permutations of the letters:**

SUMMER : 360 WATERFALL : 90,720

26
**A dog has 8 puppies, 3 male and 5 female**

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

27
**Why do you use a fundamental counting principal?**

To count the number of possibilities of the given conditions. What operation do you use for fundamental counting principals? Multiplication What is a permutation? An ordering of objects. What is the formula for nPr? What is the formula for permutations with repetition?

28
Assignment worksheet

Similar presentations

Presentation is loading. Please wait....

OK

T HE F UNDAMENTAL C OUNTING P RINCIPLE & P ERMUTATIONS.

T HE F UNDAMENTAL C OUNTING P RINCIPLE & P ERMUTATIONS.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on teaching english as a second language Download ppt on transformation of energy Ppt on pin diode circuit Ppt on different types of soil Ppt on boilers operations manual Ppt on suspension type insulators for sale Presentations ppt online viewer Maths ppt on exponents and powers Ppt on current account deficit by country Ppt on tidal power generation