Permutations and Combinations COPYRIGHT © 2006 by LAVON B. PAGE Here are the 3 cleans shirts that Ed had in the last presentation. COPYRIGHT © 2006 by LAVON B. PAGE Here are the 3 cleans shirts that Ed had in the last presentation. If he’s going to wear a different one each day on Monday, Tuesday, and Wednesday, how many choices does he have for matching up the 3 shirts with the 3 days? COPYRIGHT © 2006 by LAVON B. PAGE Monday Tuesday Wednesday COPYRIGHT © 2006 by LAVON B. PAGE Ed has 6 choices because he is arranging 3 shirts in order. The number of ways of doing this is 3 ! 2 ! 1. COPYRIGHT © 2006 by LAVON B. PAGE Ed has 6 choices because he is arranging 3 shirts in order. The number of ways of doing this is 3 ! 2 ! 1. Factorial n! = n(n-1)(n-2) . . . 1 For example, 5! = 5 ! 4 ! 3 ! 2 ! 1 = 120 Permutations A permutation of a set of objects is a listing of the objects in some specified order. The number of different permutations of n different objects is given by n!. COPYRIGHT © 2006 by LAVON B. PAGE A baseball team has nine players. How many different possible batting orders are there once it has been decided who the starting players will be? How many batting orders are possible if the pitcher is going to bat last? COPYRIGHT © 2006 by LAVON B. PAGE A baseball team has nine players. How many different possible batting orders are there once it has been decided who the starting players will be? How many batting orders are possible if the pitcher is going to bat last? There are 9 players. The number of ways to write down a batting order is 9! = 362,880 COPYRIGHT © 2006 by LAVON B. PAGE A baseball team has nine players. How many different possible batting orders are there once it has been decided who the starting players will be? How many batting orders are possible if the pitcher is going to bat last? There are 9 players. The number of ways to write down a batting order is 9! = 362,880 If the pitcher bats last, then the other 8 players can bat in any order. There are 8! = 40,320 possible arrangements. COPYRIGHT © 2006 by LAVON B. PAGE P(n,r) Given a collection of n objects, the number of different ways in which r of them can be lined up in a row is P(n,r) = n(n – 1)(n – 2) ... (n – r + 1). Such an ordered arrangement of r objects chosen from n objects is called a permutation of r objects chosen from n objects. Example: P(8,3) = 8 ! 7 ! 6 = 336 COPYRIGHT © 2006 by LAVON B. PAGE If you have five clean shirts and are going to pick one to wear on Saturday and another (different) one to wear on Sunday, how many possible ways can you make your choice? COPYRIGHT © 2006 by LAVON B. PAGE If you have five clean shirts and are going to pick one to wear on Saturday and another (different) one to wear on Sunday, how many possible ways can you make your choice? Solution: P(5,2) = 5 ! 4 = 20 COPYRIGHT © 2006 by LAVON B. PAGE If you have five clean shirts and are going to pack two of them to go on a weekend trip, how many possibilities are there for the two that you select? COPYRIGHT © 2006 by LAVON B. PAGE There are 10 possibilities for which 2 shirts you pick. The 10 ways are shown here. COPYRIGHT © 2006 by LAVON B. PAGE Combinations A set of r objects chosen from a set of n objects is called a combination of r objects chosen from n objects. The number of different combinations of r objects that may be chosen from n given objects is n! C(n, r) = r!(n ! r)! So C(n,r) represents the number of different unordered sets of r objects that could be chosen from a set of n objects. COPYRIGHT © 2006 by LAVON B. PAGE In the case of the 5 shirts where you were choosing 2 for the weekend trip, we counted 10 ways to choose 2 shirts from the 5. This is because 5! C(5, 2) = = 10 2!3! COPYRIGHT © 2006 by LAVON B. PAGE Jerry has seven compact discs that Michelle would like to borrow for a party. He has agreed to let her take four of them. In how many different ways could Michelle make her choice? COPYRIGHT © 2006 by LAVON B. PAGE Jerry has seven compact discs that Michelle would like to borrow for a party. He has agreed to let her take four of them. In how many different ways could Michelle make her choice? Michelle will choose 4 of the 7, so she can make 7! her choice in C(7,4) = = 35 ways. 4! 3! COPYRIGHT © 2006 by LAVON B. PAGE Jerry has seven compact discs that Michelle would like to borrow for a party. He has agreed to let her take four of them. In how many different ways could Michelle make her choice? Michelle will choose 4 of the 7, so she can make 7! her choice in C(7,4) = = 35 ways. 4! 3! The easy way to calculate this: 7 ! 6 ! 5 3!2 COPYRIGHT © 2006 by LAVON B. PAGE A class consists of 14 boys and 17 girls. Four students from the class are to be selected to go on a trip. (a) How many different possibilities are there for the 4 students selected to make the trip? (b) If it has been decided that 2 boys and 2 girls will make the trip, then in how many different ways could the 4 students be selected? COPYRIGHT © 2006 by LAVON B. PAGE A class consists of 14 boys and 17 girls. Four students from the class are to be selected to go on a trip. (a) How many different possibilities are there for the 4 students selected to make the trip? (b) If it has been decided that 2 boys and 2 girls will make the trip, then in how many different ways could the 4 students be selected? Solution: (a) C(31,4) = 31! = 31,465 4! 27! COPYRIGHT © 2006 by LAVON B. PAGE A class consists of 14 boys and 17 girls. Four students from the class are to be selected to go on a trip. (a) How many different possibilities are there for the 4 students selected to make the trip? (b) If it has been decided that 2 boys and 2 girls will make the trip, then in how many different ways could the 4 students be selected? Solution: 31! = 31,465 4! 27! (b) C(14,2) ! C(17,2) = 91 ! 136 = 12,376 (a) C(31,4) = COPYRIGHT © 2006 by LAVON B. PAGE A poker hand consists of 5 cards dealt from an ordinary 52-card deck. How many different poker hands are there? How many are there that give a “full house”? (A full house is a hand that contains 3 cards of one rank and 2 cards of some other rank, for example 3 aces and 2 sevens.) COPYRIGHT © 2006 by LAVON B. PAGE A poker hand consists of 5 cards dealt from an ordinary 52-card deck. How many different poker hands are there? How many are there that give a “full house”? (A full house is a hand that contains 3 cards of one rank and 2 cards of some other rank, for example 3 aces and 2 sevens.) Number of poker hands = C(52,5) = 2,598,960 COPYRIGHT © 2006 by LAVON B. PAGE A poker hand consists of 5 cards dealt from an ordinary 52-card deck. How many different poker hands are there? How many are there that give a “full house”? (A full house is a hand that contains 3 cards of one rank and 2 cards of some other rank, for example 3 aces and 2 sevens.) Number of poker hands = C(52,5) = 2,598,960 13 choices for rank of card from which 3 will be chosen, then 12 choices for rank of card from which 2 will be chosen. COPYRIGHT © 2006 by LAVON B. PAGE A poker hand consists of 5 cards dealt from an ordinary 52-card deck. How many different poker hands are there? How many are there that give a “full house”? (A full house is a hand that contains 3 cards of one rank and 2 cards of some other rank, for example 3 aces and 2 sevens.) Number of poker hands = C(52,5) = 2,598,960 13 choices for rank of card from which 3 will be chosen, then 12 choices for rank of card from which 2 will be chosen. Then choose 3 cards from the 1st rank and 2 from the 2nd rank. So answer = 13 ! 12 ! C(4,3) ! C(4,2) = 13 ! 12 ! 4 ! 6 = 3,744 COPYRIGHT © 2006 by LAVON B. PAGE Here’s a typical full house K" K# K$ 7" 7% The kings are one of the 13 ranks in the deck. So there are 13 possibilities for the rank from which 3 cards could be chosen. The 7’s are one of the remaining 12 ranks. The 3 kings shown are 3 of the 4 kings. The 2 sevens shown are 2 of the 4 sevens. Thus the number of ways of picking a full house is 13 ! 12 ! C(4,3) ! C(4,2) = 13 ! 12 ! 4 ! 6 = 3,744 COPYRIGHT © 2006 by LAVON B. PAGE How many different committees of 3 could be formed from 8 people? If Jane is one of the 8 people, how many different committees could be formed with Jane as a committee member? COPYRIGHT © 2006 by LAVON B. PAGE How many different committees of 3 could be formed from 8 people? Solution: C(8,3) = 56 possible committees COPYRIGHT © 2006 by LAVON B. PAGE How many different committees of 3 could be formed from 8 people? If Jane is one of the 8 people, how many different committees could be formed with Jane as a committee member? Solution: C(8,3) = 56 possible committees There are C(7,2) = 21 committees possible with Jane as a member. Note: You may prefer to think of this as C(1,1) ! C(7,2) COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves in the seats? — — — — — — — COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves in the seats? — — — — — — — Solution: 7! = 5040 COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves if the boys all sit together and the girls sit together? — — — — — — — COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves if the boys all sit together and the girls sit together? B B B G G G G — — — — — — — Solution: 3! ! 4! = 144 COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves if the boys all sit together and the girls sit together? B B B G G G G — — — — — — — Solution: 3! ! 4! = 144 Answer = 2 ! 144 = 288 COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves if no one sits beside a person of the same sex? — — — — — — — COPYRIGHT © 2006 by LAVON B. PAGE 3 boys and 4 girls have bought tickets for a row of 7 seats at a movie. In how many ways can they arrange themselves if no one sits beside a person of the same sex? G B G B G B G — — — — — — — Solution: 3! ! 4! = 144 COPYRIGHT © 2006 by LAVON B. PAGE

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