aim: combinations course: math lit. do now: aim: how do we determine the number of outcomes when...

Aim: Combinations Course: Math Lit.

Do Now:

Aim: How do we determine the number of outcomes when order is not an issue?

Ann, Barbara, Carol, and Dave are the only members of a school club. In how many different ways can they elect a president and treasurer for the club?

Ann, Barbara, Carol, and Dave are the only members of a school club. In how many ways can they choose 2 people to represent the club at student council meetings?

Explain how these situations are different.


Subsets & Arrangements

If order were important

{a, b}is = {b, a} ? No

A = {a, b, c, d, e}

If the two elements a and b are selected from A, then

there is one subset (order not important): {a, b}there are two arrangements (order important):

{a, b} and {b, a}

order is important

A = {Ann, Barbara, Carol, Dave}

Ann Barbara

president treasurer

Carol AnnDave CarolBarbara BarbaraDaveAnn AnnDaveCarolDaveBarbaraCarolmember of council member of council

DaveCarol


Permutation

Ann, Barbara, Carol, and Dave are the only members of a school club. In how many different ways can they elect a president and treasurer for the club?

Ann

Barbara

Carol

Dave

President

BarbaraCarolDaveAnnCarolDaveBarbaraAnnDaveBarbaraCarolAnn

Treasurer.

Ann & BarbaraAnn & CarolAnn & DaveBarbara & AnnBarbara & CarolBarbara & DaveCarol & BarbaraCarol & AnnCarol & DaveDave & BarbaraDave & CarolDave & Ann

4P2 = 4 • 3 = 12

There are 12 different arrangements of two people for president and treasurer.


Combination


Ann

Barbara

Carol

Dave

1st Person


2nd Person

Ann & BarbaraAnn & CarolAnn & DaveBarbara & AnnBarbara & CarolBarbara & DaveCarol & BarbaraCarol & AnnCarol & DaveDave & BarbaraDave & CarolDave & Ann

There are six combinations of two people that can represent


Order: Permutation vs. Combination

A selection of objects in which their order is not important.

When selecting some of the objects in the set:

The number of combinations of n objects r at a time !

n rn r

n PC

rr

!n r

n r

n PC

rr

6 36 3

6! 6! 720120(6 3)! 3! 6 20

3! 3! 3! 6 6

PC

When selecting all objects in the set: !n

PC nn

nn !n

PC nn

nn

4 44 4

4! 4! 2424(4 4)! 0! 1 1

4! 4! 4! 24 24

PC

there is only 1 combination!!

= 1= 1


Combinations

1. For any counting number n, nCn = 1

3C3 = 1 10C10 = 1

2. For any counting number n, nC0 = 1

5C0 = 1 34C0 = 1

Some Special Relationships

3. For whole numbers n and r, where r < n, nCr = nCn - r

7C3 = 7C7 - 3 = 7C4

23C16 = 23C23 - 16 = 23C7


Combinations & Pascal’s Triangle

0C0 = 1

1C0 = 1C1 = 1 1

2C0 = 2C1 = 2C2 = 1 12

3C0 = 3C1 = 3C2 = 3C3 = 1 13 3

4C0 = 4C1 = 4C2 = 4C3 = 4C4 = 11 4 46

5C0 = 5C1 = 5C2 = 5C3 = 5C4 = 5C5 = 1 155 10 10

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1


Combination


Ann

Barbara

Carol

Dave

1st Person

!r

PC rn

rn !r

PC rn

rn


2nd Person

Ann & DaveBarbara & AnnBarbara & CarolBarbara & DaveCarol & BarbaraCarol & AnnCarol & DaveDave & BarbaraDave & CarolDave & Ann

4C2 = 4P2 / 2! = 6

Ann & BarbaraAnn & Carol


Model Problems

Evaluate: 10C3 8C2

How many different three-person committees can be formed from a group of eight people? Is order important? NO

8C3 = 56

= 120 = 28

A committee has 7 men and 5 women. A subcommittee of 8 is to be formed. Write an expression for the number of ways the choice can be made.

12C8 = 495

In general, use permutations where order is important, and combinations where

order is not important.


Model Problem

Is the order of the 4 marbles important?

NO!

17 16 15 14

4 3 2 1

Combination ,!

n rn r

PC C n r

r

4 5 7= 2380

From an urn containing 4 black marbles, 8 blue marbles, and 5 red marbles, in how many ways can a set of 4 marbles be selected?

17C4 =

17 total marbles


Model Problem

If nC2 = 15, what is the value of n?

!r

PC rn

rn n P2

2!

n(n 1)

2 115

nC2

n(n - 1) = 2•15

6C2 = 15

n2 - n = 30n2 - n - 30 = 0

(n - 6)(n + 5) = 0

(n - 6) = 0 (n + 5) = 0

n = 6 n = -5


Fundamental and Combinations

A committee of five is chosen from five mathematicians and six economists. How many different committees are possible if the committee must include two mathematicians and three economists?

mathematicians:

economists:

5C2

6C3

. = 10 · 20 = 200


Model Problem

The US Senate of the 104th Congress consisted of 54 Republicans and 46 Democrats. How many committees can be formed if each committee must have 3 Republicans and 2 Democrats?

Republicans:

Democrats:

54C3

46C2

. = 24,804 · 1035

= 25,672,140


Model Problems

There are 10 boys and 20 girls in a class. Find the number of ways a team of 3 students can be selected to work on a project if the team consists of:

30C3

10C1 20C2•

20C3

20C3+10C1 20C2•

= 4060

= 10 • 190 = 1910

= 114010C0 •

= 1910 + 1140 = 3040

A. Any 3 students

B. 1 boy and 2 girls

C. 3 girls

D. At least 2 girls

2 girls

10C0 •

3 girls


Model Problem

In how many ways can 6 marbles be distributed in 3 boxes so that 3 marbles are in the first box, 2 in the second, and 1 in the third

Box 2Box 1 Box 3

6C3 3C2 1C1• •

20 3 1• •= 60


Model Problem

Find the number of ways to select 5-card hands from a standard deck so that each hand contains at most 2 aces.

at most 2 aces

Means that the hand could have 0, 1 or 2 aces

W/ no Aces 4C0 48C5

W/ 1 Aces 4C1 48C4

W/ 2 Aces 4C2 48C3

Choose Aces Complete the 5-card hand

•

•

•

= 1712304

= 778320

= 103776

= 2594400

+


Do Now:

Aim: How do we determine the number of outcomes when order is not an issue?

In the “Pick Four” Lottery, you create a 4-digit number using the numbers 1, 2, 3, 4, 5, and 6. If you play the game “straight”, you win if the winning lottery number matches your selection exactly. How many different arrangements are possible if you bet the game “straight”?

6P4 = 360


Model Problems

In the “Pick Four” Lottery, you create a 4-digit number using the numbers 1, 2, 3, 4, 5, and 6. If you play the game “straight”, you win if the winning lottery number matches your selection exactly. How many different arrangements are possible if you bet the game “straight”?

6P4 = 360

If you choose you may, you may play the game “boxed”. This means that as long as the same four numbers are chosen, regardless of order, you win. How many possible combinations are possible? 6C4 = 15


Model Problem

How many different 4-member committees can be formed from a group of 10 people if Tony, 1 of the 10 must:

A. Always on the committee

AFTER TONY IS PLACED ON THE COMMITTEE, THERE ARE 3 PLACES LEFT FOR THE OTHER 9 PEOPLE

1 • = 84 9C3

9C3 = 84

TONY IS A

MUST!


Model Problem

How many different 4-member committees can be formed from a group of 10 people if Tony, 1 of the 10 must:

B. Never be on the committee

9C4

There are now only 9 possible members for the 4-member committee

= 126


Counting Techniques

Tree Diagram Fundamental Counting Principle

Combinations Permutations

Use this to handle

inconsistencies most tedious, use when all

else fails

Counts total number of

separate tasks

Repetitions not allowed

Repetitions allowed

Subsets Arrangements

total number of ways a task

can be performed

m · n · o · p ···

Order does not matter

Order matters

!

! !n r

nC

r n r

!

!n r

nP

n r


Model Problems

Sets of 2 letters are chosen from the English alphabet. Find the number of 2-letter sets possible if the set:

a. cannot have a vowel

b. cannot have a consonant

c. must have at most 1 vowel

d. must have a vowel and a consonant


Model Problems

Find the number of ways a coach can select her starting basketball team from a group of 12 players, 8 boys and 4 girls, if the positions to be played are not taken into account, and if:

a. Sally, 1 of the players is always on the team

b. Ed, 1 of the players is never on the team.

c. both Sally and Ed are not on the team

d. either Sally or Ed, but not both, is on the team.


Model Problems

Sets of 4 letters are chosen from the English alphabet. Find the number of 4-letter lets possible if there must be the same number of vowels and consonants, and if:

a. A is always included

b. M is always included

c. E is never included

d. Q is never included

aim: combinations course: math lit. do now: aim: how do we determine the number of outcomes when...

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