MATH 105: Finite Mathematics 6-5: Combinationsmath.wallawalla.edu/~duncjo/courses/math105/.../finite_chapter_6-5.pdfDeveloping Combinations Examples of Combinations Combinations vs.

Developing Combinations Examples of Combinations Combinations vs. Permutations Conclusion

MATH 105: Finite Mathematics6-5: Combinations

Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006


Outline

1 Developing Combinations

2 Examples of Combinations

3 Combinations vs. Permutations

4 Conclusion


Outline




4 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”.


Undoing Order


Example


P(5, 5)

P(3, 3)


Undoing Order


Example


P(5, 5) ← arrange all 5 letters

P(3, 3)


Undoing Order


Example



P(3, 3) ← divide out arrangement of 3 n’s


Undoing Order


Example



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.


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.



Example


P(4, 2)

P(2, 2)

Combinations


P(n, r)

P(r , r)



Example


P(4, 2) ← arrange 2 out of 4 people

P(2, 2)

Combinations


P(n, r) ← arrange r out of n items

P(r , r)



Example



P(2, 2) ← divide out order of 2 selected people

Combinations



P(r , r) ← divide out order of r items



Example



P(2, 2) ← divide out order of 2 selected people

Combinations



P(r , r) ← divide out order of r items


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)



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)



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)



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)



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)



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)



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)



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)



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)



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)



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)



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)



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



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!


Pascal’s Triangle


Outline




4 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


Buffet Dinner

Example


C (12, 3) =12!

(12− 3)!3!=

12 · 11 · 10

3 · 2 · 1= 220



P(12, 3) = 12 · 11 · 10 = 1320


Buffet Dinner

Example


C (12, 3) =12!

(12− 3)!3!=

12 · 11 · 10

3 · 2 · 1= 220



P(12, 3) = 12 · 11 · 10 = 1320


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?


Committees

Example






Committees

Example



C (11, 3) = 165




Committees

Example



C (11, 3) = 165




Committees

Example



C (11, 3) = 165


C (6, 2) · C (5, 1) = 75



Committees

Example



C (11, 3) = 165


C (6, 2) · C (5, 1) = 75



Committees

Example



C (11, 3) = 165


C (6, 2) · C (5, 1) = 75


C (6, 2) · C (5, 1) + C (6, 3) = 95


Outline




4 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?


Club Membership

Example





Club Membership

Example



P(12, 3) = 1320



Club Membership

Example



P(12, 3) = 1320



Club Membership

Example



P(12, 3) = 1320


C (10, 3) + C (10, 3) + C (10, 4) = 450


Club Membership

Example



P(12, 3) = 1320


C (10, 3) + C (10, 3) + C (10, 4) = 450

C (12, 4)− C (10, 2) = 450


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?


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?


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)


Outline




4 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


Important Concepts



2 C (n, r) = n!(n−r)!r !




Important Concepts



2 C (n, r) = n!(n−r)!r !




Important Concepts



2 C (n, r) = n!(n−r)!r !




Important Concepts



2 C (n, r) = n!(n−r)!r !




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


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

MATH 105: Finite Mathematics 6-5: Combinationsmath.wallawalla.edu/~duncjo/courses/math105/.../finite_chapter_6-5.pdfDeveloping Combinations Examples of Combinations Combinations vs.

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