T HE F UNDAMENTAL C OUNTING P RINCIPLE & P ERMUTATIONS

Computer Science, Statistics and Probability all involve counting techniques which are a branch of mathematics called combinatorics (ways to combine things). We'll be introducing this topic in this section. For dinner you have the following choices: soupsalad chicken hamburgerprawns icecream  ENTREESMAINS DESSERTS How many different combinations of meals could you make? We'll build a tree diagram to show all of the choices.

soup salad chicken prawns hamburger chicken hamburger prawns ice cream       Now to get all possible choices we follow each path. soup, chicken, ice cream soup, chicken,  soup, prawns, ice cream soup, prawns,  soup, hamburger, ice cream soup, hamburger,  salad, chicken, ice cream salad, chicken,  salad, prawns, ice cream salad, prawns,  salad, hamburger, ice cream salad, hamburger,  Notice the number of choices at each branch 2 choices 3 choices 2 choices We ended up with 12 possibilities 2  3  2 = 12

T HE F UNDAMENTAL C OUNTING P RINCIPLE & P ERMUTATIONS E SSENTIAL Q UESTION How is the counting principle applied to determine outcomes?

Multiplication Principle of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in different ways. p  q  r  p  q  r   If we have 6 different shirts, 4 different pants, 5 different pairs of socks and 3 different pairs of shoes, how many different outfits could we wear? 6  4  5  3 = 360

T HE F UNDAMENTAL C OUNTING P RINCIPLE 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

F UNDAMENTAL C OUNTING P RINCIPLE 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

H OW 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

P HONE 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*10*10*10= 8,000,000 different numbers

T ESTING 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 = 4 10 = 1,048,576

U SING P ERMUTATIONS An ordering of n objects is a permutation of the objects.

T HERE 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

I N 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

F ACTORIAL WITH A CALCULATOR : Hit math then over, over, over. Option 4

B ACK TO THE FINALS IN THE O LYMPIC 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

P ERMUTATION OF N OBJECTS TAKEN R AT A TIME n P r =

B ACK 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! = (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