MATH 105: Finite Mathematics 6-5: Combinations Prof. Jonathan - - PowerPoint PPT Presentation

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

MATH 105: Finite Mathematics 6-5: Combinations

• Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Outline

1

Developing Combinations

2

Examples of Combinations

3

Combinations vs. Permutations

4

Conclusion

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Outline

1

Developing Combinations

2

Examples of Combinations

3

Combinations vs. Permutations

4

Conclusion

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Undoing Order

In the last section we found the number of ways to arrange the letters in the word “ninny” as follows. Example Find the number of was to arrange the letters in the word “ninny”.

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Undoing Order

In the last section we found the number of ways to arrange the letters in the word “ninny” as follows. Example Find the number of was to arrange the letters in the word “ninny”. P(5, 5) P(3, 3)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Undoing Order

In the last section we found the number of ways to arrange the letters in the word “ninny” as follows. Example Find the number of was to arrange the letters in the word “ninny”. P(5, 5) ← arrange all 5 letters P(3, 3)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Undoing Order

In the last section we found the number of ways to arrange the letters in the word “ninny” as follows. Example Find the number of was to arrange the letters in the word “ninny”. P(5, 5) ← arrange all 5 letters P(3, 3) ← divide out arrangement of 3 n’s

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Undoing Order

In the last section we found the number of ways to arrange the letters in the word “ninny” as follows. Example Find the number of was to arrange the letters in the word “ninny”. P(5, 5) ← arrange all 5 letters P(3, 3) ← 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.

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Generalizing the Concept

Example 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.

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Generalizing the Concept

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Generalizing the Concept

Example Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) ← arrange 2 out of 4 people 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) ← arrange r out of n items P(r, r)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Generalizing the Concept

Example Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) ← arrange 2 out of 4 people P(2, 2) ← 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) ← arrange r out of n items P(r, r) ← divide out order of r items

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Generalizing the Concept

Example Suppose that you want to give two movie tickets to your two closest friends. How many ways can you do this? P(4, 2) ← arrange 2 out of 4 people P(2, 2) ← 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) ← arrange r out of n items P(r, r) ← divide out order of r items

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Example Computations

Example 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!

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Pascal’s Triangle

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Outline

1

Developing Combinations

2

Examples of Combinations

3

Combinations vs. Permutations

4

Conclusion

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Buffet Dinner

Example 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) = 12 · 11 · 10 = 1320

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Buffet Dinner

Example 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! (12 − 3)!3! = 12 · 11 · 10 3 · 2 · 1 = 220 This is Not a Permutation If we had calculated using permutations, we would get: P(12, 3) = 12 · 11 · 10 = 1320

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Buffet Dinner

Example 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! (12 − 3)!3! = 12 · 11 · 10 3 · 2 · 1 = 220 This is Not a Permutation If we had calculated using permutations, we would get: P(12, 3) = 12 · 11 · 10 = 1320

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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) = 165

2 A subcommittee of 2 Republicans and 1 Democrat be chosen? 3 A subcommittee of at least 2 Republicans be chosen?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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) = 165

2 A subcommittee of 2 Republicans and 1 Democrat be chosen? 3 A subcommittee of at least 2 Republicans be chosen?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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) = 165

2 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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) = 165

2 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Committees

Example 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) = 165

2 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Outline

1

Developing Combinations

2

Examples of Combinations

3

Combinations vs. Permutations

4

Conclusion

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Club Membership

Example 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Club Membership

Example 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Club Membership

Example 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) = 1320

2 Club members David and Shauna will not work together on

the committee. How many committees are possible?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Club Membership

Example 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) = 1320

2 Club members David and Shauna will not work together on

the committee. How many committees are possible?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Club Membership

Example 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) = 1320

2 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Club Membership

Example 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) = 1320

2 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Travel Itinerary

Example 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Foot Race

Example 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?

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Poker Hands

Example 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

• f another)

3 How many hands are a full house? (3 cards of one rank, 2 of

another)

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Outline

1

Developing Combinations

2

Examples of Combinations

3

Combinations vs. Permutations

4

Conclusion

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Important Concepts

Things to Remember from Section 6-5

1 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Important Concepts

Things to Remember from Section 6-5

1 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Important Concepts

Things to Remember from Section 6-5

1 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Important Concepts

Things to Remember from Section 6-5

1 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

Important Concepts

Things to Remember from Section 6-5

1 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

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

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 365-373) Do Problem Sets 6-5 A,B

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

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 365-373) Do Problem Sets 6-5 A,B