Transcript Combinations

Aim: How do we determine the number of
outcomes when order is not an issue?
Do Now:
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: Alg. 2 & Trig.
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?
4P2 = 4 • 3 = 12
Treasurer.
President
Ann
Barbara
Carol
Dave
Barbara
Carol
Dave
Ann
Carol
Dave
Barbara
Ann
Dave
Barbara
Carol
Ann
Ann & Barbara
Ann & Carol
Ann & Dave
Barbara & Ann
Barbara & Carol
Barbara & Dave
Carol & Barbara
Carol & Ann
Carol & Dave
Dave & Barbara
Dave & Carol
Dave & Ann
Aim:are
Combinations
Course: Alg. 2 & Trig.
There
12 different arrangements
of
two people for president and treasurer.
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?
1st Person
Ann
Barbara
Carol
Dave
2nd Person
Barbara
Carol
Dave
Ann
Carol
Dave
Barbara
Ann
Dave
Barbara
Carol
Ann
Combinations
There are sixAim:combinations
of
two people that can represent
Ann & Barbara
Ann & Carol
Ann & Dave
Barbara & Ann
Barbara & Carol
Barbara & Dave
Carol & Barbara
Carol & Ann
Carol & Dave
Dave & Barbara
Dave & Carol
Dave & Ann
Course: Alg. 2 & Trig.
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  n Pr
n Cr    
 r  r!
6!
6! 720
120
(6  3)! 3!
6
6 P3




 20
6 C3 
3!
3!
3!
6
6
When selecting all
objects in the set:
Pn
n Cn 
n!
n
=1
4!
4! 24
there is only 1
P
24
(4

4)!
Aim:
Combinations
Course: Alg. 2 & Trig.
0!  1 
4 4
C




1 combination!!
4 4
4!
4!
4! 24 24
Combinations
Some Special Relationships
1. For any counting number n, nCn = 1
3C3 =
1
10C10 =
1
2. For any counting number n, nC0 = 1
5C0 =
1
34C0 =
1
3. For whole numbers n and r,
where r < n, nCr = nCn - r
7C3
= 7C7 - 3 = 7C4
23C16 = 23C23 - 16
Aim: Combinations
=
23C7
Course: Alg. 2 & Trig.
Probability Notation
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?
1st Person
2nd Person
Ann
Pr

n C r Barbara
r!
n
Carol
Dave
Aim: Combinations
Barbara
Carol
Dave
Ann
Carol
Dave
Barbara
Ann
Dave
Barbara
Carol
Ann
4C2
Ann & Barbara
Ann & Carol
Ann & Dave
Barbara & Ann
Barbara & Carol
Barbara & Dave
Carol & Barbara
Carol & Ann
Carol & Dave
Dave & Barbara
Dave & Carol
Dave & Ann
Course: Alg. 2 & Trig.
= 4P2 / 2! = 6
Model Problems
Evaluate: 10C3 = 120
8C2 = 28
How many different three-person
committees can be formed from a group of
eight people?
Is order important? NO
8C3
= 56
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
Aim: Combinations
Course: Alg. 2 & Trig.
order is not important.
Model Problem
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?
Is the order of the 4 marbles important?
NO!
Combination
Pr
n Cr 
r!
n
17 total marbles
4
17C4 =
5 7
17  16  15  14
4 3  2 1
Aim: Combinations
= 2380
Course: Alg. 2 & Trig.
Model Problem
If nC2 = 15, what is the value of n?
Pr
n Cr 
r!
n(n  1)
n P2



15
C
n 2
2!
21
n
n(n - 1) = 2•15
n2 - n = 30
n2 - n - 30 = 0
(n - 6)(n + 5) = 0
(n - 6) = 0
n=6
6C2
Aim: Combinations
(n + 5) = 0
n = -5
= 15
Course: Alg. 2 & Trig.
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:
A. Any 3 students
30C3
B. 1 boy and 2 girls
10C1
= 4060
•
20C2
= 10 • 190 = 1900
C. 3 girls
10C0
•
20C3
2 girls
D. At least 2 girls
Aim: Combinations
10C1
•
20C2
= 1140
3 girls
+
10C0
•
20C3
= 1900 + 1140
= 3040
Course: Alg.
2 & Trig.
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 1
6C3
20
Box 2
•
•
Aim: Combinations
3C2
3
Box 3
•
•
1C1
1
= 60
Course: Alg. 2 & Trig.
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
Course: Alg. 2 & Trig.
possible? Aim: Combinations6C4 = 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:
9C3 = 84
A. Always on the committee
1 • 9C3
TONY
IS A
MUST!
Aim: Combinations
= 84
AFTER TONY IS PLACED ON
THE COMMITTEE, THERE
ARE 3 PLACES LEFT FOR
THE OTHER 9 PEOPLE
Course: Alg. 2 & Trig.
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
There are now only 9 possible
members for the 4-member committee
9C4
Aim: Combinations
= 126
Course: Alg. 2 & Trig.
Regents Problem
The principal would like to assemble a
committee of 8 students from the 15-member
student council. How many different
committees can be chosen?
(1) 120
(2) 6,435
Aim: Combinations
(3) 32,432,400
(4) 259,459,200
Course: Alg. 2 & Trig.
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
= 1712304
W/ 1 Aces
4C1
•
48C4
= 778320
W/ 2 Aces
4C2
•
48C3
+ = 103776
Aim: Combinations Complete the
Choose Aces
5-card hand
= 2594400
Course: Alg. 2 & Trig.
Probability Involving Combinations
Two cards are drawn at random from a standard
deck of 52 cards, without replacement. What is the
probability that both cards drawn are fives?
Dependent Events
1
2
3
4C2
Counting
Principle
P(1st 5) = 4/52 P(2d 5) = 3/51
P(5, 5) = 4/52 • 3/51
Permutations
Combinations
Number of
ways to draw
2Aim:
5’sCombinations
from
possible 4
4P2
52P2
4C2
52C2
52C2
= 1/221
= 1/221
Number of
ways to draw
any
cards
Course: 2
Alg.
2 & Trig.
from deck of 52
Model Problem
In a school organization, there are 4
sophomores and 5 juniors. A committee of 4
people is to be selected from this group.
What is the probability that 2 sophomores
and 2 juniors will be on the committee?
n( E ) Permutation
P ( E) 
n( S )
or
Combination
Pr
n Cr 
r!
n
P(2 sophomores, 2 juniors on
the 4-member committee)
=
n(combinations of 2s & 2j)
n(4-member
combinations madeCourse:
from
Aim: Combinations
Alg. 2 & Trig.
of the 9 members)
Model Problem
In a school organization, there are 4
sophomores and 5 juniors. A committee of 4
people is to be selected from this group.
What is the probability that 2 sophomores
and 2 juniors will be on the committee?
Total Outcomes n(S):
Nine total students from
: 9C4 = 126
which to choose
Successful Outcomes n(E):
How do we determine n(2 soph., 2 jun.)?
Two of four sophomores
4C2 = 6
Two of five juniors
5C2 = 10
n(2 soph.,
2 jun.) = 4C2 • 5C2Course:
= 6•10
= 60
Aim: Combinations
Alg. 2 & Trig.
Model Problem
In a school organization, there are 4
sophomores and 5 juniors. A committee of 4
people is to be selected from this group.
What is the probability that 2 sophomores
and 2 juniors will be on the committee?
P(2 sophs., 2 juniors) =
4C2 • 5C2
9C4
= 6 • 10
126
= 60
126
Aim: Combinations
= 10
21
Course: Alg. 2 & Trig.
Model Problem
An urn contains 4 white marbles and 5 blue
marbles, all of equal size. Three marbles are
drawn at random with no replacement.
What is the probability that at least 2
marbles drawn are blue?
Total Outcomes n(S) :
9C3
Successful Outcomes n(E)
At least 2 are blue: (2-b, 1-w) or (3-b)
P(A  B)
= P(A) + P(B) - P(A  B)
5C2 • 4C1
5C3
Successful Outcomes n(E) = 5C2 • 4C1 + 5C3
Total Outcomes n(S)
9C3
Aim: Combinations
40 + 10Course: Alg. 2 & Trig.
P(at least 2 b) =
= 25/42
Model Problem
A candy dish contains 10 candies. Three
candies are covered with red foil and 7 with
green foil.
A. If 2 candies are chosen at random from
the dish, what is the probability that both
will be covered with the same colored foil?
P(both red or both green) = P(2R) + P(2G)
P(A  B) = P(A) + P(B) - P(A  B)
P(2R) = 3C2
P(2G) = 7C2
10C2
10C2
P(2R or 2G) =
3C2
Aim: Combinations
+ 7C2 = 3 + 21 = 8/15
45
10C2
Course: Alg. 2 & Trig.
Model Problem
A candy dish contains 10 candies. Three
candies are covered with red foil and 7 with
green foil.
B. If 2 candies are chosen at random from
the dish, what is the probability that each
will be covered with a different foil?
Aim: Combinations
Course: Alg. 2 & Trig.
Model Problem
A candy dish contains 10 candies. Three
candies are covered with red foil and 7 with
green foil.
C. If 5 candies are chosen at random from
the dish, what is the probability that at least
3 will be covered with green foil?
Aim: Combinations
Course: Alg. 2 & Trig.