slide 7- 1 copyright © 2006 pearson education, inc. publishing as pearson addison-wesley

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Sequences, Series, and Combinatorics

Chapter 10


10.1Sequences and Series

Find terms of sequences given the nth term.

Look for a pattern in the sequence and try to determine a general term.

Convert between sigma notation and other notation for a series.

Construct the terms of a recursively defined sequence.


Sequences

A sequence is a function, where the domain is a set of consecutive positive integers beginning with 1.

An infinite sequence is a function having for its domain the set of positive integers, {1, 2, 3, 4, 5, …}.

A finite sequence is a function having for its domain a set of positive integers, {1, 2, 3, 4, 5, …, n}, for some positive integer n.


Sequence Formulas

In a formula, the function values are known as terms of the sequence. The first term in a sequence is denoted as a1,

the third term as a3 ,

and the nth term, or the general term, as an.


Example

Find the first 4 terms and the 9th term of the sequence whose general term is given by an = 4(2)n.

Solution: We have an = 4(2)n, so

a1 = 4(2)1 = 8 a2 = 4(2)2 = 16

a3 = 4(2)3 = 32

a4 = 4(2)4 = 64

a9 = 4(2)9 = 2048

The power (2)n causes the sign of the terms to alternate between positive and negative, depending on whether the n is even or odd. This kind of sequence is called an alternating sequence.


Another Example

Example: Predict the general term of the sequence

2, 4, 8, 16, …

Solution: These are the powers of 2, so the general

term might be 2(2)n1.


Sums and Series


Example

For the sequence 1, 3, 5, 7, 9, 11, 13, … find each of the following:

a) S1 b) S5 c) S7

Solution: S1 = 1 S5 = 1 + 3 + (5) + 7 + (9) = 5 S7 = 1 + 3 + (5) + 7 + (9) + 11 + (13) = 7


Sigma Notation

The Greek letter (sigma) can be used to simplify notation when the general term of a sequence is a formula. For example, the sum of the first three terms ofthe sequence ,…, ,… can be named as follows, using sigma notation, or summationnotation: . This is read “the sum as k goes

from 1 to 3 of .” The letter k is called the index of summation. The index of summation might be a number other than 1, and a letter other than k can be used.

1 92

2 2 21

2k

32

1

1

2k

k

21

2k


More Examples

Find and evaluate the sum.

Solution: (1)232+(1)333 + (1)434

= 9 + (27) + 81 = 63

Write sigma notation for the sum 5 + 25 + 125 + … Solution: This in an infinite series, so we use the infinity symbol to write the

sigma notation.

4

2

( 1) 3k k

k

1

5k

k


Recursive Definitions

A sequence may be defined recursively or by using a recursion formula. Such a definition lists the first term, or the first few terms, and then describes how to determine the remaining terms from the given terms.

Example: Find the first 4 terms of the sequence defined by

a1 = 3, an+1= 3an 2 for n 1.

Solution:

a1 = 3

a2 = 3a1 2 = 3 3 2 = 7

a3 = 3a2 2 = 3 7 2 = 19

a4 = 3a3 2 = 3 19 2 = 55


10.2Arithmetic Sequences

and Series For any arithmetic sequence, find the nth term when n is

given and n when the nth term is given, and given two terms, find the common difference and construct the sequence.

Find the sum of the first n terms of an arithmetic sequence.


Arithmetic Sequences

A sequence in which each term after the first is found by adding the same number to the preceding term is an arithmetic sequence.

A sequence is arithmetic if there exists a number d, called the common difference, such that an+1 = an + d for any integer n 1.


Example

For each of the following arithmetic sequences, identify the first term, a1, and the common difference, d.

6, 10, 14, 18, 22, … 0, 6, 12, 18, 24, …

Solution: The first term a1 is the first term listed. To find the common difference, d, we choose any term beyond the first and subtract the preceding term from it.

1 2 4 5, ,1, , ,...

3 3 3 3


Example continued

We obtained the common difference by subtracting a1 from a2. Had we subtracted a2 from a3 or a3 from a4, we would have obtained the same values for d. We can check by adding d to each term in a sequence to see if we progress correctly to the next term.

First Term, a1

1

3 c) .

6 (6 0 = 6)0b) 0, 6, 12, 18,24, …

4 (10 6 = 4)6a) 6, 10, 14, 18, 22, …

Common Difference, dSequence

1 2 4 5, ,1, , ,...

3 3 3 3

2 1 1( )3 3 3

1

3


nth Term of an Arithmetic Sequence

To find a formula for the general, or nth, term of any arithmetic sequence, we denote the common difference by d, write out the first few terms, and look for a pattern.

The nth term of an arithmetic sequence is given by the formula: an = a1 + (n 1)d, for any integer .1n


Example

Find the 11th term of the arithmetic sequence

2, 6, 10, 14, …

Solution: We first note that a1 = 2, d = 4, and n = 11. Then using the formula for the nth term, we obtain

an = a1 + (n 1)d

a11 = 2 + (11 1)4

a11 = 2 + 40

a11 = 42

The 11th term is 42.


Another Example

The 3rd term of an arithmetic sequence is 5, and the 9th term is 37. Find a1 and d and construct the sequence.

Solution: We know that a3 = 5 and a9 = 37. Thus we have to add d 6 times to get 37 from 5.

5 + 6d = 37 6d = 42 d = 7

Since a3 = 5, we subtract d twice to get a1.

a1= 5 2(7) = 19 The sequence is 19, 12, 5, 2, …

In general, d should be subtracted n 1 times from an in order to find a1.


Sum of the First n Terms

The formula for the sum of the first n terms of an arithmetic sequence is given by:

( ).2n 1 n

nS = a +a


Example

Find the sum of the first 11 terms of the arithmetic sequence 16, 12, 8, 4, …

Solution: Note that a1 = 16, d = 4, and n = 11. First we find the last term a11.

a11 = 16 + (11 1)(4) = 16 40 = 24

Thus,

The sum of the first 11 terms is 44.

11

11 11(16 24) ( 8) 44

2 2S


Another Example

Find the sum: .

Solution: It is helpful to write out a few terms first:

14 + 24 + 34 + .

It appears that a1 = 14, d = 10, n = 10.

We then find the last term. a10 = 10(10) + 4

= 104

Thus,

10

1

(10 4)k

k

10

10(14 104)

25(118)

590.

S


Application

An orchestra consists of 8 rows of musicians. The first row has 5 musicians, the second row has 7 musicians, and the third row has 9 musicians.

How many musicians are in the last row? What is the total number of musicians in the

orchestra?

Solution:

We need to find a8 to find the number of musicians in the last row. a8 = 5 + (8 1)2

a8 = 19 There are 19 musicians in the last row.


Application continued

We can then use the formula to find the total number of musicians.

There are a total of 96 musicians in the orchestra.

( )2n 1 n

nS = a +a

1

8

8

8

( )28

(5 19)24(24)

96

n n

nS a a

S

S

S


10.3Geometric Sequences

and Series Identify the common ratio of a geometric sequence,

and find a given term and the sum of the first n terms.

Find the sum of an infinite geometric series, if it exists.


Geometric Sequences

A sequence in which each term after the first is found by multiplying the preceding term by the same number is a geometric sequence.

The number that is multiplied by each term to produce the next term is called the common ratio.

A sequence is geometric if there is a number r, called the common ratio, such that or an+1 = anr for any integer n 1.

1n

n

ar

a


Example

For each of the following geometric sequences, identify the common ratio.

4, 16, 64, 256, 1024, …

, …

$750, $600, $480, $381, …

Solution: To find the common ratio, we divide any term (other than the first) by the preceding term.

2 4 8 161, , , ,

3 9 27 81


Example continued

0.8 ( , and so on)

c) $750, $600, $480, $381, …

,

and so on)

, …

4 ( and so on)a) 4, 16, 64, 256, 1024, …

Common RatioSequence

2 4 8 161, , , ,

3 9 27 81

16 644, 4,

4 16

2 42 23 9( ,

21 3 33

2

3

600 4800.8, 0.8

750 600


nth Term of a Geometric Sequence

The nth term of a geometric sequence is given by the formula: an = a1r n 1, for any integer .

Example: Find the 11th term of the geometric sequence 1, 3, 9, 27, …

Solution: We first note that a1 = 1, and n = 11. We then find the common ratio or 3.

Then using the formula an = a1r n 1, we have

The 11th term is 59,049.

1n

31

r

11 1 10

11 1 3 1 3 1 59,049 59,049a


Sum of the First n Terms

The formula for the sum of the first n terms of a geometric sequence is given by:

, for any r 1.

Example: Find the sum:

Solution: This is a geometric series with a1 = 6, r = 3, and n = 12.

1(1 )

1

n

n

a rS

r

12

1

2(3 ) k

k

12

12

6(1 3 )

1 31,594,320

S


Infinite Geometric Series

The sum of the terms of an infinite geometric sequence is an infinite geometric series. For some geometric sequences, Sn gets close to a specific number as n gets large. Consider the infinite series

+ … .

Each of the partial sums is less than 1, but Sn gets very close to 1 as n gets large.

1 1 1 1 1

2 4 8 16 2n


Infinite Geometric Series continued

We say that 1 is the limit of Sn and also that 1 is the sum of the infinite geometric sequence. The sum of an infinite geometric sequence is denoted S . In this case S = 1.


Limit or Sum of an Infinite Geometric Series

When , the limit or sum of an infinite geometric series is given by

Examples: Determine whether each of the following infinite geometric series has a limit. If a limit exists, find it.

3 + 6 + 12 + 24 +

1r 1

1

aS

r

4 4 44

3 9 27


Solutions

Here r = 2, so . Since , the series does not have a limit.

Here , so . Since , the series

does have a limit. We find the limit:

2 2 r 1r

1

3r 1 1

3 3r 1r

1 4 43

1 41 1 ( )3 3

aS

r


Application

A racquetball hit in the air 27 feet rebounds to two-thirds of its previous height after each bounce. Find the total vertical distance the ball has traveled when it hits the ground the tenth time.

Between bounces, the racquetball travels twice its rebound height. The geometric series for the total distance traveled is 54 + 36 + 24 + .


Application continued

Using a1 = 54 and r = :

The ball has traveled about 159.2 feet by the tenth

bounce.

2

3

1

10

10

(1 )

12

54(1 ( ) )32

13

159.2

n

n

a rS

r

S


10.4Mathematical Induction

List the statements of an infinite sequence that is defined by a formula.

Do proofs by mathematical induction.


Sequences of Statements

Infinite sequences of statements occur often in mathematics. In an infinite sequence of statements, there is a statement for each natural number.

Example: List the four statements in the sequence that can be obtained from each of the following.

log n < n 1 + 3 + 5 + + (2n 1) = n2


Solutions

This time, Sn is “log n < n.”

S1: log 1 < 1

S2: log 2 < 2

S3: log 3 < 3

S4: log 4 < 4

This time, Sn is “1 + 3 + 5 + + (2n 1) = n2.”

S1: 1 = 12

S2: 1 + 3 = 22

S3: 1 + 3 + 5 = 32

S4: 1 + 3 + 5 + 7 = 42


Proving Infinite Sequences of Statements

Mathematical induction can be used to try to prove that all statements in an infinite sequence of statements are true.

The statements usually have the form:

“For all natural numbers n, Sn”, where Sn is some mathematical sentence.


The Principle of Mathematical Induction

We can prove an infinite sequence of statements Sn by showing the following.

Basis step. S1 is true. Induction step. For all natural numbers k, Sk Sk+1.

When you are learning to do proofs by mathematical induction, it is helpful to first write out Sn , S1, Sk , and Sk + 1. This helps to identify what is to be assumed and what is to be deduced.


Example

Prove: For every natural number n,

1 + 3 + 5 + + (2n 1) = n2.

Proof: We first list Sn , S1, Sk , and Sk+1.

Sn: 1 + 3 + 5 + + (2n 1) = n2

S1: 1 = 12

Sk: 1 + 3 + 5 + + (2k 1) = k2

Sk + 1: 1 + 3 + 5 + + (2k 1) + [2(k + 1) 1] = (k + 1)2

Basis step. S1, as listed, is true. Induction step. We let k be any natural number. We assume

Sk to be true and try to show that it implies that Sk+1 is true.


Example continued

Now, Sk is 1 + 3 + 5 + + (2k 1) = k2.

Starting with the left side of Sk+1 and substituting k2 for1 + 3 + 5 + + (2k 1), we have 1 + 3 + + (2k 1) + [2(k + 1) 1]

= k2 + [2(k + 1) 1] = k2 + 2k + 1 = (k + 1)2.

We have derived Sk+1 from Sk. Thus we have shown that for all

natural numbers k, Sk Sk+1. The basis step and the induction step tell us that the proof is complete.


Another Example

Prove: For every natural number n, n < 2n.

Proof: We first list Sn , S1, Sk , and Sk+1.

Sn: n < 2n

S1: 1 < 21

Sk: k < 2k

Sk + 1: k + 1 < 2k+1

Basis step. S1, as listed, is true since 21= 2 and 1 < 2. Induction step. We let k be any natural number. We

assume Sk to be true and try to show that it implies that Sk+1 is true.


Another Example continued

Now,

k < 2k This is Sk. 2k < 2 2k Multiplying by 2 on both sides

2k < 2k+1 Adding exponents on the right

k + k < 2k+1 Rewriting 2k as k + k

Since k is any natural number, we now that 1 k. Thus, k + 1 k + k. Adding k on both sides

Putting the results k + 1 k + k and k + k < 2k+1 together gives us k + 1 < 2k+1 . This is Sk+1.

We have derived Sk+1 from Sk. Thus we have shown that for all natural numbers k, Sk Sk+1. The basis step and the induction step tell us that the proof is complete.


10.5Combinatorics: Permutations

Evaluate factorial and permutation notation and solve related applied problems.


Permutations

The study of permutations involves order and arrangements.

Example: How many 3-letter code symbols can be formed with the letters A, B, C without repetition (that is, using each letter only once)?

Solution: Consider placing the letters in these boxes.

We can select any of the 3 letters for the first letter in the symbol. Once this letter has been selected, the second must be selected from the 2 remaining letters. After this, the third letter is determined, since only 1 possibility is left.


Permutations continued

The possibilities can also be arrived at by using a tree diagram, as shown.


The Fundamental Counting Principle

Given a combined action, or event, in which the first action can be performed in n1 ways, the second action can be performed in n2 ways, and so on, the total number of ways in which the combined action can be performed is the product

Outcomes — the results of an experiment Event — set of outcomes

1 2 3 ... . kn n n n


Example

How many 4-letter code symbols can be formed with the letters A, B, C, D, E, and F with repetition (that is, allowing letters to be repeated)?

Solution: Since repetition is allowed, there are 6 choices for the first, second, third, and

fourth letter. Thus, by the fundamental counting principle, there are 6 • 6 • 6 • 6, or 1296 code symbols.


Definitions

A permutation of a set of n objects is an ordered arrangement of all n objects.

The Total Number of Permutations of n Objects

The total number of permutations of n objects, denoted

nPn, is given by

( 1)( 2)... 3 2 1. n nP n n n


Example

Find each of the following.

a) 3P3 b) 5P5

Solution:

a) 3P3 = 3 • 2 • 1 = 6

b) 5P5 = 5 • 4 • 3 • 2 • 1 = 120

Start with 3.

3 factors


Factorial Notation

For any natural number n,n! = n(n 1)(n 2) … 3 • 2 • 1.

For the number 0, 0! = 1.

nPn = n!

For any natural number n, n! = n(n 1)!.


Example

Rewrite 8! with a factor of 6!.

Solution: 8! = 8 • 7! = 8 • 7 • 6!

A permutation of a set of n objects taken k at a time is an ordered arrangement of k objects taken from the set.


The Number of Permutations of n Objects Taken k at a Time

The number of permutations of a set of n objects taken k at a time, denoted nPk, is given by

factors

( 1)( 2)... [ ( 1)] (1)

!(2)

( )!

n k

k

P n n n n k

n

n k


Example

Compute 7P3 using both forms of the formula.

Form 1:

Form 2:

The 7 tells where to start.

37 6 57 210 P

The 3 tells how many factors.

37

!

( )!

7!

4!

7 6 5 4! 7 6 5 4!

4

7

7

!

3

P

4!

210


Another Example

How many different ways can 8 people be seated in a row of 5 chairs?

Solution: We are determining the number of permutations of 8 objects taken 5 at a time.

There is no repetition of people.

Using form (2), we get 8 5

8!

(8 5)!

8!6720

3!

P


Permutations of Sets with Nondistinguishable Objects

For a set of n objects in which n1 are of one kind, n2 are of another kind, …, and nk are of a kth kind, the number of distinguishable permutations is

1 2

!

! ! ... !k

n

n n n


Example

In how many distinguishable ways can the letters of the word MISSISSIPPI be arranged?

Solution: There are 4 I’s, 4 S’s, 2 P’s, and 1 M for a total of 11 letters. Thus,

11!34,650

4! 4! 2! 1!


10.6Combinatorics: Combinations

Evaluate combination notation and solve related applied problems.


Combinations

We sometimes make a selection from a set without regard to order. Such a selection is called a combination.

Example: Find all the combinations of 2 letters taken from a set of 3 letters {A, B, C}.

Solution: The combinations are {A, B}, {A, C}, and{B, C}. There are 3 combinations of the 3 letters taken 2 at a time.


Definitions

Subset: Set A is a subset of set B, denoted A B, if every element of A is an element of B.

Combination: A combination containing k objects is a subset containing k objects.

Combination Notation: The number of combinations of n objects taken k at a time is denoted nCk.


Combinations of n Objects Taken k at a Time

The total number of combinations of n objects taken k at a time, denoted nCk, is given by

!(1)

!( )!

( 1)( 2)...[ ( 1)](2)

! !

n k

n kn k

nC

k n k

or

P n n n n kC

k k


Binomial Coefficient Notation

Another kind of notation for nCk is binomial coefficient notation.

n k

nC

k


Example

Evaluate , using forms

(1) and (2). Solution: a) By form (1),

b) By form (2),8

3

8 8! 8!

3 3!(8 3)! 3!5!

8 7 6 5 4 3 2 1 8 756

3 2 1 5 4 3 2 1 1

8 8 7 6 8 756

3 3 2 1 1

The 8 tells where to start.

The 3 tells how many factors there are in both the numerator and the denominator and where to start in the denominator.


Subsets of Size k and of Size n k

The number of subsets of size k of a set with n objects is the same as the number of subsets of size n k. The number of combinations of n objects taken k at a time is the same as the number of combinations of n objects taken n k at a time.

and n k n n k

n nC C

k n k


Example

A team manager has 11 students who are qualified to play basketball. How many different 5-person teams can be chosen?

Solution:

There are 462 different 5-person teams that could be chosen.

11 5

11 11! 11!462

5 5!(11 5)! 5!6!C


Another Example

How many committees can be formed from a group of 8 seniors and 10 juniors if each committee consists of 5 seniors and 6 juniors?

Solution: The 5 seniors can be selected in 8C5 ways and the 6 juniors can be selected in 10C6 ways.

If we use the fundamental counting principle, it follows that the number of possible committees is

8 5 10 6

8! 10!

5!3! 6!4!8 7 6 5! 10 9 8 7 6!

5! 3 2 1 6! 4 3 2 1

8 7 6

C C

5!

5! 3 2 10 9 8

1

7 6!

6! 4 3 2 1

56 63011,760

1 3


10.7The Binomial Theorem

Expand a power of a binomial using Pascal’s triangle or factorial notation.

Find a specific term of a binomial expansion.

Find the total number of subsets of a set of n objects.


Pascal’s Triangle


The Binomial Theorem Using Pascal’s Triangle


Example

Expand (u v)4.

Solution: We have (a + b)n, where a = u, b = v, and n = 4. We use the 5th row of Pascal’s Triangle:

1 4 6 4 1

Then we have:4 4 3 1 2 2 1 3 4

4 3 2 2 3 4

1 4 6( ) ( ) 4( ) ( ) ( ) ( ) ( ) 1( ) ( )

4 6 4

u v u u v u v u v v

u u v u v uv v


Another Example

Expand (x 3y)4. a = x, b = 3y, and n = 4. We use the 5th row of Pascal’s

triangle: 1 4 6 4 1

Then we have

4 3 1 2 2 3 4

4 3 2 2 3 4

( ) ( ) ( 3 ) ( ) ( 3 ) ( )( 3 ) ( 3 )

12 54 1

1 4 6 4

08 8

1

1

x x y x y x y y

x x y x y xy y


The Binomial Theorem Using Factorial Notation


Example


Finding a Specific Term

Finding the (k + 1)st Term

The (k + 1)st term of (a + b)n is

Example: Find the 7th term in the expansion (x2 2y)11.

First, we note that 7 = 6 + 1. Thus, k = 6, a = x2, b = 2y, and n = 11. Then the 7th term of the expansion is

.n k kna b

k

11 6 56 62 2 10 611 11!2 or 2 , or 29,568

6 6!5!x y x y x y


Total Number of Subsets

The total number of subsets of a set with n elements is 2n.

Example: The set {A, B, C, D, E, F} has how many subsets?

Solution: The set has 6 elements, so the number of subsets is 26 or 64.


10.8Probability

Compute the probability of a simple event.


Probabilities

Experimental probabilities are determined by making observations and gathering data.

Theoretical probabilities are determined by reasoning mathematically.

Principle P (Experimental)

Given an experiment in which n observations are made, if a situation, or event, E occurs m times out of n observations, then we say that the experimental probability of the event, P(E), is given by ( ) .

mP E

n


Example

Sociological Survey. The authors of this text conducted an experimental survey to determine the number of people who are left-handed, right-handed, or both. The results are shown in the graph.

a) Determine the probability that a person is right-handed.

b) Determine the probability that a person is left-handed.


Example continued

c) Determine the probability that a person is ambidextrous (uses both hands with equal ability).

d) There are 120 bowlers in most tournaments held by the Professional Bowlers Association. On the basis of the data, how many of the bowlers would you expect to be left-handed?

Solutions: a) The number of right-

handed is 82, the number of left-handed is 17, the number of ambidextrous is 1. The total number of observations is 82 + 17 + 1 = 100.

The probability that a person is right-handed is

82, or 0.82, or 82%

100P


Example continued

b) The probability that a person is left-handed is P, where

c) The probability that a person is ambidextrous is P, where

d) There are 120 bowlers, and from part (b) we can expect 17% to be left-handed. Since 17% of 120 is 0.17(120) = 20.4, we can expect that about 20 of the bowlers will be left-handed.

17, or 0.17, or 17%

100P

1, or 0.1, or 1%

100P


Theoretical Probability

Each possible result of an experiment, such as flipping a coin, throwing a dart, or choosing a card from a deck, is called an outcome.

The set of all possible outcomes is called the sample space.

An event is a set of outcomes, that is, a subset of the sample space.


Example

Consider this dartboard. Assume that the experiment is “throwing a dart” and that the dart hits the board. Find each of the following.

a) The outcomes b) The sample space

Solution: a) The outcomes are hitting

white (W), purple (P), or yellow (Y).

b) The sample space is {hitting white, hitting purple, hitting yellow}, which can be stated as {W, P, Y}.


Principle P (Theoretical)

If an event E can occur m ways out of n possible equally likely outcomes of a sample space S, then the theoretical probability of the event, P(E), is given by

Example: What is the probability of rolling a 4 on a die?

Solution: On a fair die, there are 6 equally likely outcomes and there is 1 way to roll a 4.

( ) .m

P En

16(4)P


Examples

What is the probability of rolling an odd number?

Solution: The event is rolling an odd number. It can occur 3 ways (getting a 1, 3, or 5). The number of equally likely outcomes is 6.

Suppose we select, without looking, one marble from a bag containing 4 blue and 9 purple marbles. What is the probability of selecting a blue marble?

Solution: There are 13 equally likely ways of selecting any marble, and 4 ways of selecting blue.

3 1(odd)

6 2P

(selecting a blue marble3

) = 4

1P


Probability Properties

a) If an event E cannot occur, then P(E) = 0.

b) If an event E is certain to occur, then P(E) = 1.

c) The probability that an event E will occur is a number from 0 to 1: 0 P(E) 1.


Example

Suppose that 4 people are selected at random from a group of 7 men and 5 women. What is the probability that 2 men and 2 women are selected?

Solution: The number of ways of selecting 4 people from a group of 12 is 12C4. Two men can be selected in 7C2 ways, and 2 women can be selected in 5C2 ways. By the fundamental counting principle, the number of ways of selecting 2 men and 2 women is 7C2 • 5C2.

7 2 5 2

12 4

210 14

495 33

C CP

C


Another Example

What is the probability of getting a total of 5 on a roll of a pair of dice?

Solution: On each die, there are 6 possible outcomes. The outcomes are paired so there are 6(6) or 36 possible ways in which the two can fall.

There are 4 ways to roll a total of 5. (1, 4) (4, 1) (2, 3) and (3, 2)

4 1(5)

36 9P

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