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Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion
MATH 105: Finite Mathematics6-5: 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 theletters 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 theletters 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 theletters 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 theletters 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 theletters 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 toundoing 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 twoclosest friends. How many ways can you do this?
Combinations
A combination of n things taken r at a time is the number of waysto select r things from n distinct things without replacement whenthe 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 twoclosest 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 waysto select r things from n distinct things without replacement whenthe 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 twoclosest 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 waysto select r things from n distinct things without replacement whenthe 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 twoclosest 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 waysto select r things from n distinct things without replacement whenthe 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 twoclosest 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 waysto select r things from n distinct things without replacement whenthe 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 thesalad bar, you choose 3 of them. In how many ways can you makethis 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 thesalad bar, you choose 3 of them. In how many ways can you makethis 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 thesalad bar, you choose 3 of them. In how many ways can you makethis 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 5Democrats. 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 5Democrats. 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 5Democrats. 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 5Democrats. 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 5Democrats. 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 5Democrats. 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 5Democrats. 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 holdmore than one position?
2 Club members David and Shauna will not work together onthe 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 holdmore than one position?
2 Club members David and Shauna will not work together onthe 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 holdmore than one position?
P(12, 3) = 1320
2 Club members David and Shauna will not work together onthe 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 holdmore than one position?
P(12, 3) = 1320
2 Club members David and Shauna will not work together onthe 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 holdmore than one position?
P(12, 3) = 1320
2 Club members David and Shauna will not work together onthe 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 holdmore than one position?
P(12, 3) = 1320
2 Club members David and Shauna will not work together onthe 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 citiesshe 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, andBronze medals are awarded to first, second and third placerespectively. 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 amedal?
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 thesuits are red: hearts and diamonds; two are black: spades andclubs. 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, 1of another)
3 How many hands are a full house? (3 cards of one rank, 2 ofanother)
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 ournew-found skills at counting to determine the probability orlikelihood 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 ournew-found skills at counting to determine the probability orlikelihood of a given event.
For next time
Read Section 7-1 (pp 365-373)
Do Problem Sets 6-5 A,B