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Combinations and Permutations #1 - J. T. Butler
1
1Comb’s and Permu’s -1 J. T. Butler
Rule of Sum andRule of Sum andProductProduct
Rule of sum:
If one event can occur in m ways andanother event can occur in n ways,there are m + n ways in which exactlyone event can occur.
2Comb’s and Permu’s -1 J. T. Butler
Example
lSuppose there are 5 red ballsand 10 green balls. Then,there are 5+10=15 ways tochoose a ball of any color.
3Comb’s and Permu’s -1 J. T. Butler
Rule of Product:
If one event can occur in mways and another event canoccur in n ways, then there arem . n ways the two events canoccur together.
4Comb’s and Permu’s -1 J. T. Butler
ExamplelSuppose there are 5 red balls
and 10 green balls. Then,there are 5 . 10 = 50 ways tochoose a red ball AND agreen ball.
5Comb’s and Permu’s -1 J. T. Butler
ExampleThe address register in your PC has 24bits. How many locations can yourPC access?
Each bit can be chosen independentlyin 2 ways. By the rule of product,there are 2 . 2 .... . 2=2n = 16,777,216or 16 megabytes.
6Comb’s and Permu’s -1 J. T. Butler
PermutationsPermutationsHow many ways P(n,n) are there to arrangen of n objects?
By the rule of product
P (n,n) = n × (n −1) × (n −2) × ... × 2 × 1
n ways to n −1 ways n −2 wayschoose to choose to choose1st object 2nd object 3rd object
Combinations and Permutations #1 - J. T. Butler
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7Comb’s and Permu’s -1 J. T. Butler
n × (n-1) × (n-2) × ... × 2 × 1 = n!
“n factorial”
Also,
This is called permutation.
P n r n n n n r
nn r
( , ) ( ) ( ) ( )
!
( )!
= × − × − × × − +
=−
1 2 1K
8Comb’s and Permu’s -1 J. T. Butler
P (n,r) = n (n −1) (n −2) ... (n −r +1) can beviewed as a distribution of balls to cells
r positions (cells)6444447444448
n ways to n −1 ways to n −r +1 waysfill the fill the fill the1st position 2nd position rth position
L
9Comb’s and Permu’s -1 J. T. Butler
Notation:
( , ) is the number of permutations of
objects from a set of distinct objects
Alternative notation:
( ( , ))
P n r r
n
P P P n rrn
n r= =
10Comb’s and Permu’s -1 J. T. Butler
RememberRememberGiven a set S of n distinct objects, apermutation is an arrangement orordering of objects from S.
11Comb’s and Permu’s -1 J. T. Butler
How many permutations are there of twoelements of S={a,b,c,d}?
There are 12: ab, ba, ac, ca, ad, da, bc,cb, bd, db, cd, and dc.
P(4,2) = 4 . 3 = 12.
ExampleExample
12Comb’s and Permu’s -1 J. T. Butler
How many ways are there to arrange the three
letters in ALL?
If all letters were distinct,like AL L , there
be 3! = 6. But two of letters are identical.
ALL AL L AL L
LAL L AL L AL
LLA L L A L L A
1 2
1 2 2 1
1 2 2 1
1 2 2 1
⇔⇔⇔
ExampleExample
Combinations and Permutations #1 - J. T. Butler
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13Comb’s and Permu’s -1 J. T. Butler
We observe that :
(the number of arrangements
of ALL) = number of arrange -
ments of AL L That is, the
number of arrangements of ALL
= 3!
2!
1 2
2
3
!
.
.
×
=
14Comb’s and Permu’s -1 J. T. Butler
In general,If there are objects of the first type,
objects of the second type, ... , objects
of the - th type, such that the total number
of objects , then the
number of arrangements of the objects is
!
n n
n
r
n n n n n
n
nn n n
r
r
r
1 2
1 2
1 2
= + + +...
! !... !.
15Comb’s and Permu’s -1 J. T. Butler
!
is called a
nn n n
multinomialr1 2! !... !
.
16Comb’s and Permu’s -1 J. T. Butler
ExampleIf six people, A, B, C, D, E, and F are seated about around table, how many circular arrangements arepossible?
C
C
D
AB
E
F
A
D
FE
B
D
A
CB
F
E
A
B
EF
C
D
(a) (b) (c) (d)Here, (a) and (b) are identical whereas (a)=(b), (c), and(d) are different.
17Comb’s and Permu’s -1 J. T. Butler
We can arbitrarily choose one person,say A, to be at the top. Then, we can choose the order of the remainingones in 5! ways. Thus, the number ofcircular permutations on 6 objects is 5!.
18Comb’s and Permu’s -1 J. T. Butler
Suppose that A, B, and C are females, and
D, E, and F are males. We now want to
arrange them so the sexes alternate. How
many ways are there to place the people?
Choose A, a female to be in top position.
There are 3! ways to arrange the males and 2!
ways to arrange the remaining females or
3! 2! = 12 ways.×
Combinations and Permutations #1 - J. T. Butler
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19Comb’s and Permu’s -1 J. T. Butler
CombinationsCombinationsHow many ways are there to arrange5 red balls and 10 green balls?
There are 15 objects, 5 of the first kindand 10 of the second. Thus, there are
15!5! 10!
arrangements.
20Comb’s and Permu’s -1 J. T. Butler
Suppose we want the number of ways
to 5 objects from 15 distinct
objects. Notice, there is a one - to - one
correspondence between an arrangement
of 5 red balls and 10 green balls and a
selection of 5 objects from 15. Let the
red balls signify which object is selected.
choose
21Comb’s and Permu’s -1 J. T. Butler
Let be the number of ways to
select objects from . Then, we saw
In general,
C n r
r n
C C n r
nr n r
P n rr
( , )
( , )!
!. ( , )
!
!( )!
( , )
!.
15 515
5 10!= =
−=
22Comb’s and Permu’s -1 J. T. Butler
Alternative words:
C(n,r) is the number of ways to choose robjects from n distinct objects.
C(n,r) <=> “n choose r”
Alternative notation:
C Cn
rC n rr
nn r= =
FHGIKJ = ( , )
23Comb’s and Permu’s -1 J. T. Butler
Notes:
1. =!
!( - )!=
-.
2. When one chooses an object, it cannot
be chosen again. This is choice
repetition.
n
rn
r n r
n
n r
without
FHGIKJ
FHG
IKJ
24Comb’s and Permu’s -1 J. T. Butler
NoteNoteGiven a set S of n distinct objects, acombination is a subset S’ of S.combination <=> selection
Given a set S of n distinct objects, apermutation is an arrangement of elements.permutation <=> arrangement
Combinations and Permutations #1 - J. T. Butler
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25Comb’s and Permu’s -1 J. T. Butler
Question:Let S = {a,b,c,d}. Then {a,b} and {c,d}are combinations. How manycombinations of size 2 are there of S?
Answer:{a,b}, {a,c}, {a,d}, {b,c}, {b,d}, and{c,d} are all of the combinations of S ofsize 2. There are 6. There are 12permutations, two for each combinationabove.
26Comb’s and Permu’s -1 J. T. Butler
Rule of sum:C(n,r) = C(n−1, r) + C(n−1, r−1)
Choose r objects Choose the special object from n− 1 objects and r − 1 other objects not including a from n − 1 objects not special object. including the special
object.
27Comb’s and Permu’s -1 J. T. Butler
C(4,2) = C(3,2) +C(3,1) = 3+3 = 6
6447448 6447448 {b,c}, {b,d}, {c,d} {a,b}, {a,c}, {a,d}
a is the special object.
28Comb’s and Permu’s -1 J. T. Butler
Pascal’s TrianglePascal’s TriangleSum Diff.
1 1 1 1 1 2 0 1 2 1 4 0 1 3 3 1 8 0 1 4 6 4 1 16 0 1 5 10 10 5 1 32 0 1 6 15 20 15 6 1 64 0
29Comb’s and Permu’s -1 J. T. Butler
How to build Pascal’s triangle
1. Write the 1’s 1 1 1 1 1
2. Fill in the remaining numbers by adding pairs 1
1 1
1 2 1
1 3 3 130Comb’s and Permu’s -1 J. T. Butler
Brief historyThis was first published in 1653by Pascal in one of the firstpublications on probability.
Combinations and Permutations #1 - J. T. Butler
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31Comb’s and Permu’s -1 J. T. Butler
r
n 0 Sum Diff.
0 1 1 1 11 2 1 1 2 02 3 1 2 1 4 03 4 1 3 3 1 8 04 5 1 4 6 4 1 16 05 6 1 5 10 10 5 1 32 06 7 1 6 15 20 15 6 1 64 0
32Comb’s and Permu’s -1 J. T. Butler
Note:Pascal’s Rule is used to derive theentries of this table. That is,C(0,0) = 1
and
C (n,r) = C (n −1, r) +C (n −1, r −1).
33Comb’s and Permu’s -1 J. T. Butler
r
n 0 Sum Diff.
0 1 1 1 11 2 1 +1 2 02 3 1 +2 +1 4 03 4 1 +3 +3 +1 8 04 5 1 +4 +6 +4 +1 16 05 6 1 +5 +10 +10 +5 +1 32 06 7 1 +6 +15 +20 +15 +6 +1 64 0
34Comb’s and Permu’s -1 J. T. Butler
r
n 0 Sum Diff.0 1 1 1 11 2 1 -1 2 02 3 1 -2 +1 4 03 4 1 -3 +3 -1 8 04 5 1 -4 +6 -4 +1 16 05 6 1 -5 +10 -10 +5 -1 32 06 7 1 -6 +15 -20 +15 -6 +1 64 0
35Comb’s and Permu’s -1 J. T. Butler
Note:C(n, r) is the number of ways to chooser objects from n distinct objects.
C(n, r) is called a “binomial coeffi-cient”.
Binomial coefficients satisfy thousandsof identities.
36Comb’s and Permu’s -1 J. T. Butler
Naming the AceNaming the AceParadoxParadox
Combinations and Permutations #1 - J. T. Butler
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37Comb’s and Permu’s -1 J. T. Butler
A Deck of Cards is a set of 52 cards
Each card as a value
2 3 4 5 6 7 8 9 10 J Q K Awhere J = Jack, Q = Queen K = King, and A = Ace
and a suit
Hearts Diamonds Spades Clubs
38Comb’s and Permu’s -1 J. T. Butler
EXAMPLE:
There are four aces
ace of heartsace of diamondsace of spadesace of clubs
Similarly, there are four kings, fourqueens, etc.
39Comb’s and Permu’s -1 J. T. Butler
Hearts
Diamonds
Spades
Clubs
40Comb’s and Permu’s -1 J. T. Butler
Definition:“To deal 52 cards to 4 people”
means to randomly give 13 cards to 4people. Usually, a person knowsonly those cards given to him/her.
Definition:A hand is a set of 13 cards given
to one person.
41Comb’s and Permu’s -1 J. T. Butler
Question: Naming the ace
52 cards are distributed to 4 people. Oneperson says “I have an ace.” What is theprobability he/she has another ace?
Answer: 5359/14498 < .
However, suppose he/she says, “I have anace of spades.” What is the probabilityhe/she has another ace?
Answer: 11686/20825 > !
1
2
1
242Comb’s and Permu’s -1 J. T. Butler
Question:
Why should naming the ace changethe probability?
Combinations and Permutations #1 - J. T. Butler
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43Comb’s and Permu’s -1 J. T. Butler
Small DeckSmall DeckShown below is a much smaller deck.In this case, we give two cards eachto two people.
44Comb’s and Permu’s -1 J. T. Butler
Below are the six different hands withtwo cards
1
2
3
4
5
6
45Comb’s and Permu’s -1 J. T. Butler
There are 6 ways a person can have ahand of two cards, as shown. Ifsomeone says, “I have an ace”, theprobability he/she has another ace is1/5 (there are 5 hands with at least oneace, only one of which has anotherace). If he/she says, “I have an ace ofspades,” then the probability he/shehas another ace is 1/3.
46Comb’s and Permu’s -1 J. T. Butler
Mr. Smith has two children. At leastone is a boy. What is the probabilitythat the other is a boy also?
Children Paradox
47Comb’s and Permu’s -1 J. T. Butler
There are four equally likely outcomes
Older Younger
Boy BoyBoy GirlGirl BoyGirl Girl (impossible)
Since only one of three possibleoutcomes consists of two boys, theprobability is 1/3 !
48Comb’s and Permu’s -1 J. T. Butler
Playing KenoPlaying KenoQuestion: Keno is a game in which youchoose 6 distinct numbers from 1 to 80.The casino chooses 20 distinct numbersfrom 1 to 80. You enter a choice ofnumbers for $0.60 and the casino paysyou an amount of money depending onhow many of the 6 numbers you havechosen matches the numbers chosen bythe casino.
Combinations and Permutations #1 - J. T. Butler
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49Comb’s and Permu’s -1 J. T. Butler
For example, you win $1,000 if all
six numbers you have chosen were
also chosen by the casino. What is
the probability ( ) that exactly of
the numbers you have chosen were
chosen by the casino, for ?
P k k
k0 6≤ ≤
50Comb’s and Permu’s -1 J. T. Butler
Answer:Number of Number of Number ofways k of 80 ways casino ways playernumbers can can choose can choosematch remaining remaining
numbers numbers
Total number of Total number of ways casino can ways player can choose numbers choose numbers
P kk
k
k k( ) =
FHG
IKJ
−
−FHG
IKJ −FHG
IKJ
FHG
IKJFHG
IKJ
80 80
20
60
680
20
80
6
(1)
51Comb’s and Permu’s -1 J. T. Butler
k P(k) Class exp.
0 0.167 0.1381 0.363 0.2782 0.308 0.3053 0.130 0.2504 0.029 0.0285 0.003 0.0006 0.0001 0.000
This table shows the exactprobability, as calculatedfrom (1). Also shown isthe probability as extrac-ted during a class experi-ment in which 40 studentschose 6 numbers and theinstructor chose 20.
52Comb’s and Permu’s -1 J. T. Butler
ExactExperiment
00.10.20.30.4
0 1 2 3 4 5 6
00.10.20.30.4
ProbabilityProbability
k
Playing KenoProbability of matching k numbersProbability of matching k numbers
53Comb’s and Permu’s -1 J. T. Butler
How to Become RichHow to Become Richby Playing Kenoby Playing Keno
You cannot.Spend yourmoney some-where else.
54Comb’s and Permu’s -1 J. T. Butler