MATH 105: Finite Mathematics 6-5: Combinations
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1 MATH 105: Finite Mathematics 6-5: Combinations Prof. Jonathan Duncan Walla Walla College Winter Quarter, 2006
2 Outline 1 Developing Combinations 2 s of Combinations 3 Combinations vs. Permutations 4 Conclusion
3 Outline 1 Developing Combinations 2 s of Combinations 3 Combinations vs. Permutations 4 Conclusion
4 Undoing Order In the last section we found the number of ways to arrange the letters in the word ninny as follows. Find the number of was to arrange the letters in the word ninny.
5 Undoing Order In the last section we found the number of ways to arrange the letters in the word ninny as follows. Find the number of was to arrange the letters in the word ninny. P(5, 5) P(3, 3)
6 Undoing Order In the last section we found the number of ways to arrange the letters in the word ninny as follows. Find the number of was to arrange the letters in the word ninny. P(5, 5) P(3, 3) arrange all 5 letters
7 Undoing Order In the last section we found the number of ways to arrange the letters in the word ninny as follows. Find the number of was to arrange the letters in the word ninny. P(5, 5) P(3, 3) arrange all 5 letters divide out arrangement of 3 n s
8 Undoing Order In the last section we found the number of ways to arrange the letters in the word ninny as follows. Find the number of was to arrange the letters in the word ninny. P(5, 5) P(3, 3) arrange all 5 letters divide out arrangement of 3 n s Dividing out the order of the n s is something we can generalize to undoing the order of selection all together.
9 Generalizing the Concept Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? Combinations A combination of n things taken r at a time is the number of ways to select r things from n distinct things without replacement when the order of selection does not matter.
10 Generalizing the Concept Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) P(2, 2) Combinations A combination of n things taken r at a time is the number of ways to select r things from n distinct things without replacement when the order of selection does not matter. P(n, r) P(r, r)
11 Generalizing the Concept Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) P(2, 2) arrange 2 out of 4 people Combinations A combination of n things taken r at a time is the number of ways to select r things from n distinct things without replacement when the order of selection does not matter. P(n, r) P(r, r) arrange r out of n items
12 Generalizing the Concept Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) P(2, 2) arrange 2 out of 4 people divide out order of 2 selected people Combinations A combination of n things taken r at a time is the number of ways to select r things from n distinct things without replacement when the order of selection does not matter. P(n, r) P(r, r) arrange r out of n items divide out order of r items
13 Generalizing the Concept Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) P(2, 2) arrange 2 out of 4 people divide out order of 2 selected people Combinations A combination of n things taken r at a time is the number of ways to select r things from n distinct things without replacement when the order of selection does not matter. P(n, r) P(r, r) arrange r out of n items divide out order of r items
14 Computations Find each value 1 C(5, 0) 2 C(5, 1) 3 C(5, 2) 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
15 Computations Find each value 1 C(5, 0) 2 C(5, 1) 3 C(5, 2) 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
16 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) 3 C(5, 2) 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
17 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) 3 C(5, 2) 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
18 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
19 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
20 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
21 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) 5 C(5, 4) 6 C(5, 5)
22 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) = 5! (5 3)!3! = 5! 2!3! = 10 5 C(5, 4) 6 C(5, 5)
23 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) = 5! (5 3)!3! = 5! 2!3! = 10 5 C(5, 4) 6 C(5, 5)
24 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) = 5! (5 3)!3! = 5! 2!3! = 10 5 C(5, 4) = 5! (5 4)!4! = 5! 1!4! = 5 6 C(5, 5)
25 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) = 5! (5 3)!3! = 5! 2!3! = 10 5 C(5, 4) = 5! (5 4)!4! = 5! 1!4! = 5 6 C(5, 5)
26 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) = 5! (5 3)!3! = 5! 2!3! = 10 5 C(5, 4) = 5! (5 4)!4! = 5! 1!4! = 5 6 C(5, 5) = 5! (5 5)!5! = 5! 0!5! = 1
27 Computations Find each value 1 C(5, 0) = 5! (5 0)!0! = 5! 5!0! = 1 2 C(5, 1) = 5! (5 1)!1! = 5! 4!1! = 5 3 C(5, 2) = 5! (5 2)!2! = 5! 3!2! = 10 4 C(5, 3) = 5! (5 3)!3! = 5! 2!3! = 10 5 C(5, 4) = 5! (5 4)!4! = 5! 1!4! = 5 6 C(5, 5) = 5! (5 5)!5! = 5! 0!5! = 1 Symmetric!
28 Pascal s Triangle
29 Outline 1 Developing Combinations 2 s of Combinations 3 Combinations vs. Permutations 4 Conclusion
30 Buffet Dinner A buffet dinner offers 12 different salads. On your first trip to the salad bar, you choose 3 of them. In how many ways can you make this choice? This is Not a Permutation If we had calculated using permutations, we would get: P(12, 3) = = 1320
31 Buffet Dinner A buffet dinner offers 12 different salads. On your first trip to the salad bar, you choose 3 of them. In how many ways can you make this choice? C(12, 3) = 12! = = 220 (12 3)!3! This is Not a Permutation If we had calculated using permutations, we would get: P(12, 3) = = 1320
32 Buffet Dinner A buffet dinner offers 12 different salads. On your first trip to the salad bar, you choose 3 of them. In how many ways can you make this choice? C(12, 3) = 12! = = 220 (12 3)!3! This is Not a Permutation If we had calculated using permutations, we would get: P(12, 3) = = 1320
33 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? 2 A subcommittee of 2 Republicans and 1 Democrat be chosen? 3 A subcommittee of at least 2 Republicans be chosen?
34 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? 2 A subcommittee of 2 Republicans and 1 Democrat be chosen? 3 A subcommittee of at least 2 Republicans be chosen?
35 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? C(11, 3) = A subcommittee of 2 Republicans and 1 Democrat be chosen? 3 A subcommittee of at least 2 Republicans be chosen?
36 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? C(11, 3) = A subcommittee of 2 Republicans and 1 Democrat be chosen? 3 A subcommittee of at least 2 Republicans be chosen?
37 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? C(11, 3) = A subcommittee of 2 Republicans and 1 Democrat be chosen? C(6, 2) C(5, 1) = 75 3 A subcommittee of at least 2 Republicans be chosen?
38 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? C(11, 3) = A subcommittee of 2 Republicans and 1 Democrat be chosen? C(6, 2) C(5, 1) = 75 3 A subcommittee of at least 2 Republicans be chosen?
39 Committees A congressional committee consists of 6 Republicans and 5 Democrats. In how many ways can: 1 A subcommittee of 3 representatives be chosen? C(11, 3) = A subcommittee of 2 Republicans and 1 Democrat be chosen? C(6, 2) C(5, 1) = 75 3 A subcommittee of at least 2 Republicans be chosen? C(6, 2) C(5, 1) + C(6, 3) = 95
40 Outline 1 Developing Combinations 2 s of Combinations 3 Combinations vs. Permutations 4 Conclusion
41 Club Membership A club with 12 members wishes to elect a president, vice-president, and treasurer and to choose a 4 member activities committee. 1 In how many ways can officers be elected if no one may hold more than one position? 2 Club members David and Shauna will not work together on the committee. How many committees are possible?
42 Club Membership A club with 12 members wishes to elect a president, vice-president, and treasurer and to choose a 4 member activities committee. 1 In how many ways can officers be elected if no one may hold more than one position? 2 Club members David and Shauna will not work together on the committee. How many committees are possible?
43 Club Membership A club with 12 members wishes to elect a president, vice-president, and treasurer and to choose a 4 member activities committee. 1 In how many ways can officers be elected if no one may hold more than one position? P(12, 3) = Club members David and Shauna will not work together on the committee. How many committees are possible?
44 Club Membership A club with 12 members wishes to elect a president, vice-president, and treasurer and to choose a 4 member activities committee. 1 In how many ways can officers be elected if no one may hold more than one position? P(12, 3) = Club members David and Shauna will not work together on the committee. How many committees are possible?
45 Club Membership A club with 12 members wishes to elect a president, vice-president, and treasurer and to choose a 4 member activities committee. 1 In how many ways can officers be elected if no one may hold more than one position? P(12, 3) = Club members David and Shauna will not work together on the committee. How many committees are possible? C(10, 3) + C(10, 3) + C(10, 4) = 450
46 Club Membership A club with 12 members wishes to elect a president, vice-president, and treasurer and to choose a 4 member activities committee. 1 In how many ways can officers be elected if no one may hold more than one position? P(12, 3) = Club members David and Shauna will not work together on the committee. How many committees are possible? C(10, 3) + C(10, 3) + C(10, 4) = 450 C(12, 4) C(10, 2) = 450
47 Travel Itinerary A traveler wishes to visit 3 of Amsterdam, Barcelona, Copenhagen, Rome, and Zurich on her trip. An itinerary is a list of the 3 cities she will visit. 1 How many itineraries are possible? 2 How many include Copenhagen as the first stop? 3 How many include Copenhagen as any stop? 4 How many include Copenhagen and Rome?
48 Foot Race Ten people participate in a foot race in which Gold, Silver, and Bronze medals are awarded to first, second and third place respectively. Bob and Carol both participate in the race. 1 How many ways can the medals be awarded? 2 How many ways can the medals be awarded if Bob wins a medal? 3 In how many ways can Bob and Carol finish consequitively?
49 Poker Hands A deck of playing cards consists of 52 cards in 4 suits. Two of the suits are red: hearts and diamonds; two are black: spades and clubs. In each suit, there are 13 ranks: 2, 3, 4,..., 10, J, Q, K, A. In a typical Poker hand, 5 cards are dealt. 1 How many different poker hands are possible? 2 How many hands are four of a kind? (4 cards of one rank, 1 of another) 3 How many hands are a full house? (3 cards of one rank, 2 of another)
50 Outline 1 Developing Combinations 2 s of Combinations 3 Combinations vs. Permutations 4 Conclusion
51 Important Concepts Things to Remember from Section Combinations are used when order is not important 2 C(n, r) = n! (n r)!r! 3 Pascal s Triangle 4 Differentiating between Combinations and Permutations
52 Important Concepts Things to Remember from Section Combinations are used when order is not important 2 C(n, r) = n! (n r)!r! 3 Pascal s Triangle 4 Differentiating between Combinations and Permutations
53 Important Concepts Things to Remember from Section Combinations are used when order is not important 2 C(n, r) = n! (n r)!r! 3 Pascal s Triangle 4 Differentiating between Combinations and Permutations
54 Important Concepts Things to Remember from Section Combinations are used when order is not important 2 C(n, r) = n! (n r)!r! 3 Pascal s Triangle 4 Differentiating between Combinations and Permutations
55 Important Concepts Things to Remember from Section Combinations are used when order is not important 2 C(n, r) = n! (n r)!r! 3 Pascal s Triangle 4 Differentiating between Combinations and Permutations
56 Next Time... Chapter 7 starts next time. In chapter 7 we will apply our new-found skills at counting to determine the probability or likelihood of a given event. For next time Read Section 7-1 (pp ) Do Problem Sets 6-5 A,B
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