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.
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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.
Aim: Combinations Course: Math Lit.
Combination
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?
Ann
Barbara
Carol
Dave
1st Person
BarbaraCarolDaveAnnCarolDaveBarbaraAnnDaveBarbaraCarolAnn
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
Combination
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?
Ann
Barbara
Carol
Dave
1st Person
!r
PC rn
rn !r
PC rn
rn
BarbaraCarolDaveAnnCarolDaveBarbaraAnnDaveBarbaraCarolAnn
2nd Person
Ann & DaveBarbara & AnnBarbara & CarolBarbara & DaveCarol & BarbaraCarol & AnnCarol & DaveDave & BarbaraDave & CarolDave & Ann
4C2 = 4P2 / 2! = 6
Ann & BarbaraAnn & Carol
Aim: Combinations Course: Math Lit.
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.
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
+
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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!
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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
Aim: Combinations Course: Math Lit.
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.
Aim: Combinations Course: Math Lit.
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