THE FUNDAMENTAL COUNTING PRINCIPLE PERMUTATIONS THE FUNDAMENTAL COUNTING

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 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

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

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 Fundamental Counting Principle 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

FACTORIAL WITH A CALCULATOR: • Hit math then over, over. • Option 4

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