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Math 1300: Section 7- 4 Permutations and Combinations
1. Factorials
Permutations
Combinations
Math 1300 Finite Mathematics
Section 7-4: Permutations and Combinations
Jason Aubrey
Department of Mathematics
University of Missouri
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Jason Aubrey Math 1300 Finite Mathematics
2. Factorials
Permutations
Combinations
Problem 1: Consider the set {p, e, n}. How many two-letter
"words" (including nonsense words) can be formed from the
members of this set, if two different letters have to be used?
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Jason Aubrey Math 1300 Finite Mathematics
3. Factorials
Permutations
Combinations
Problem 1: Consider the set {p, e, n}. How many two-letter
"words" (including nonsense words) can be formed from the
members of this set, if two different letters have to be used?
We list all possibilities: pe, pn, en, ep, np, ne, a total of 6.
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Jason Aubrey Math 1300 Finite Mathematics
4. Factorials
Permutations
Combinations
Problem 1: Consider the set {p, e, n}. How many two-letter
"words" (including nonsense words) can be formed from the
members of this set, if two different letters have to be used?
We list all possibilities: pe, pn, en, ep, np, ne, a total of 6.
Problem 2: Now consider the set consisting of three males:
{Paul, Ed, Nick }. For simplicity, denote the set by {p, e, n}.
How many two-man crews can be selected from this set?
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Jason Aubrey Math 1300 Finite Mathematics
5. Factorials
Permutations
Combinations
Problem 1: Consider the set {p, e, n}. How many two-letter
"words" (including nonsense words) can be formed from the
members of this set, if two different letters have to be used?
We list all possibilities: pe, pn, en, ep, np, ne, a total of 6.
Problem 2: Now consider the set consisting of three males:
{Paul, Ed, Nick }. For simplicity, denote the set by {p, e, n}.
How many two-man crews can be selected from this set?
pe (Paul, Ed), pn (Paul, Nick) and en (Ed, Nick)
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Jason Aubrey Math 1300 Finite Mathematics
6. Factorials
Permutations
Combinations
Both problems involved counting the numbers of arrangements
of the same set {p, e, n}, taken 2 elements at a time, without
allowing repetition.
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Jason Aubrey Math 1300 Finite Mathematics
7. Factorials
Permutations
Combinations
Both problems involved counting the numbers of arrangements
of the same set {p, e, n}, taken 2 elements at a time, without
allowing repetition. However, in the first problem, the order of
the arrangements mattered since pe and ep are two different
"words".
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Jason Aubrey Math 1300 Finite Mathematics
8. Factorials
Permutations
Combinations
Both problems involved counting the numbers of arrangements
of the same set {p, e, n}, taken 2 elements at a time, without
allowing repetition. However, in the first problem, the order of
the arrangements mattered since pe and ep are two different
"words". In the second problem, the order did not matter since
pe and ep represented the same two-man crew. We counted
this only once.
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Jason Aubrey Math 1300 Finite Mathematics
9. Factorials
Permutations
Combinations
Both problems involved counting the numbers of arrangements
of the same set {p, e, n}, taken 2 elements at a time, without
allowing repetition. However, in the first problem, the order of
the arrangements mattered since pe and ep are two different
"words". In the second problem, the order did not matter since
pe and ep represented the same two-man crew. We counted
this only once. The first example was concerned with counting
the number of permutations of 3 objects taken 2 at a time.
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Jason Aubrey Math 1300 Finite Mathematics
10. Factorials
Permutations
Combinations
Both problems involved counting the numbers of arrangements
of the same set {p, e, n}, taken 2 elements at a time, without
allowing repetition. However, in the first problem, the order of
the arrangements mattered since pe and ep are two different
"words". In the second problem, the order did not matter since
pe and ep represented the same two-man crew. We counted
this only once. The first example was concerned with counting
the number of permutations of 3 objects taken 2 at a time. The
second example was concerned with the number of
combinations of 3 objects taken 2 at a time.
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Jason Aubrey Math 1300 Finite Mathematics
11. Factorials
Permutations
Combinations
Definition (Factorial)
For n a natural number,
n! = n(n − 1)(n − 2) · · · 2 · 1
0! = 1
n! = n · (n − 1)!
Note: Many calculators have an n! key or its equivalent
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Jason Aubrey Math 1300 Finite Mathematics
12. Factorials
Permutations
Combinations
Examples
3! = 3(2)(1) = 6
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Jason Aubrey Math 1300 Finite Mathematics
17. Factorials
Permutations
Combinations
Definition (Permutation of a Set of Objects)
A permutation of a set of distinct objects is an arrangement of
the objects in a specific order without repetition.
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Jason Aubrey Math 1300 Finite Mathematics
18. Factorials
Permutations
Combinations
Theorem (Number of Permutations of n Objects)
The number of permutations of n distinct objects without
repetition, denoted by Pn,n is
Pn,n = n(n − 1) · · · 2 · 1 = n!
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Jason Aubrey Math 1300 Finite Mathematics
19. Factorials
Permutations
Combinations
Example: Now suppose that the director of the art gallery
decides to use only 2 of the 4 available paintings, and they will
be arranged on the wall from left to right. We are now talking
about a particular arrangement of 2 paintings out of the 4,
which is called a permutation of 4 objects taken 2 at a time.
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Jason Aubrey Math 1300 Finite Mathematics
20. Factorials
Permutations
Combinations
Example: Now suppose that the director of the art gallery
decides to use only 2 of the 4 available paintings, and they will
be arranged on the wall from left to right. We are now talking
about a particular arrangement of 2 paintings out of the 4,
which is called a permutation of 4 objects taken 2 at a time.
×
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Jason Aubrey Math 1300 Finite Mathematics
21. Factorials
Permutations
Combinations
Example: Now suppose that the director of the art gallery
decides to use only 2 of the 4 available paintings, and they will
be arranged on the wall from left to right. We are now talking
about a particular arrangement of 2 paintings out of the 4,
which is called a permutation of 4 objects taken 2 at a time.
4×
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Jason Aubrey Math 1300 Finite Mathematics
22. Factorials
Permutations
Combinations
Example: Now suppose that the director of the art gallery
decides to use only 2 of the 4 available paintings, and they will
be arranged on the wall from left to right. We are now talking
about a particular arrangement of 2 paintings out of the 4,
which is called a permutation of 4 objects taken 2 at a time.
4×3
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Jason Aubrey Math 1300 Finite Mathematics
23. Factorials
Permutations
Combinations
Example: Now suppose that the director of the art gallery
decides to use only 2 of the 4 available paintings, and they will
be arranged on the wall from left to right. We are now talking
about a particular arrangement of 2 paintings out of the 4,
which is called a permutation of 4 objects taken 2 at a time.
4 × 3 = 12
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Jason Aubrey Math 1300 Finite Mathematics
24. Factorials
Permutations
Combinations
Definition (Permutation of n Objects Taken r at a Time)
A permutation of a set of n distinct objects taken r at a time
without repetition is an arrangement of r of the n objects in a
specific order.
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Jason Aubrey Math 1300 Finite Mathematics
25. Factorials
Permutations
Combinations
Theorem (Number of Permutations of n Objects Taken r at a
Time)
The number of permutations of n distinct objects taken r at a
time without repetition is given by
Pn,r = n(n − 1)(n − 2) · · · (n − r + 1)
or
n!
Pn,r = 0≤r ≤n
(n − r )!
n! n!
Note: Pn,n = (n−n)! = 0! = n! permutations of n objects taken n
at a time.
Note: In place of Pn,r the symbol P(n, r ) is often used.
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Jason Aubrey Math 1300 Finite Mathematics
26. Factorials
Permutations
Combinations
Example: Find P(5, 3)
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Jason Aubrey Math 1300 Finite Mathematics
28. Factorials
Permutations
Combinations
Example: Find P(5, 3)
P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60.
This means there are 60 permutations of 5 items taken 3 at a
time.
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Jason Aubrey Math 1300 Finite Mathematics
29. Factorials
Permutations
Combinations
Example: Find P(5, 3)
P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60.
This means there are 60 permutations of 5 items taken 3 at a
time.
Example: Find P(5, 5), the number of permutations of 5
objects taken 5 at a time.
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Jason Aubrey Math 1300 Finite Mathematics
30. Factorials
Permutations
Combinations
Example: Find P(5, 3)
P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60.
This means there are 60 permutations of 5 items taken 3 at a
time.
Example: Find P(5, 5), the number of permutations of 5
objects taken 5 at a time.
P(5, 5) = 5(4)(3)(2)(1) = 120.
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Jason Aubrey Math 1300 Finite Mathematics
31. Factorials
Permutations
Combinations
Example: A park bench can seat 3 people. How many seating
arrangements are possible if 3 people out of a group of 5 sit
down?
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Jason Aubrey Math 1300 Finite Mathematics
32. Factorials
Permutations
Combinations
Example: A park bench can seat 3 people. How many seating
arrangements are possible if 3 people out of a group of 5 sit
down?
P(5, 3) = (5)(4)(3) = 60.
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Jason Aubrey Math 1300 Finite Mathematics
33. Factorials
Permutations
Combinations
Example: A bookshelf has space for exactly 5 books. How
many different ways can 5 books be arranged on this
bookshelf?
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Jason Aubrey Math 1300 Finite Mathematics
34. Factorials
Permutations
Combinations
Example: A bookshelf has space for exactly 5 books. How
many different ways can 5 books be arranged on this
bookshelf?
P(5, 5) = 5(4)(3)(2)(1) = 120
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Jason Aubrey Math 1300 Finite Mathematics
35. Factorials
Permutations
Combinations
Definition (Combination of n Objects Taken r at a Time)
A combination of a set of n distinct objects taken r at a time
without repetition is an r -element subset of the set of n objects.
The arrangement of the elements in the subset does not matter.
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Jason Aubrey Math 1300 Finite Mathematics
36. Factorials
Permutations
Combinations
Theorem (Number of Combinations of n Objects Taken r at a
Time)
The number of combinations of n distinct objects taken r at a
time without repetition is given by:
n
Cn,r =
r
Pn,r
=
r!
n!
= 0≤r ≤n
r !(n − r )!
n
Note: In place of the symbols Cn,r and , the symbols
r university-logo
C(n, r ) is often used.
Jason Aubrey Math 1300 Finite Mathematics
39. Factorials
Permutations
Combinations
Example: In how many ways can you choose 5 out of 10
friends to invite to a dinner party?
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Jason Aubrey Math 1300 Finite Mathematics
40. Factorials
Permutations
Combinations
Example: In how many ways can you choose 5 out of 10
friends to invite to a dinner party?
Does the order of selection matter?
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Jason Aubrey Math 1300 Finite Mathematics
41. Factorials
Permutations
Combinations
Example: In how many ways can you choose 5 out of 10
friends to invite to a dinner party?
Does the order of selection matter?
No! So we use combinations. . .
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Jason Aubrey Math 1300 Finite Mathematics
42. Factorials
Permutations
Combinations
Example: In how many ways can you choose 5 out of 10
friends to invite to a dinner party?
Does the order of selection matter?
No! So we use combinations. . .
P(10, 5) 10(9)(8)(7)(6)
C(10, 5) = = = 252
5! 5(4)(3)(2)(1)
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Jason Aubrey Math 1300 Finite Mathematics
43. Factorials
Permutations
Combinations
Example: How many 5-card poker hands will have 3 aces and
2 kings?
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Jason Aubrey Math 1300 Finite Mathematics
44. Factorials
Permutations
Combinations
Example: How many 5-card poker hands will have 3 aces and
2 kings?
The solution involves both the multiplication principle and
combinations.
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Jason Aubrey Math 1300 Finite Mathematics
45. Factorials
Permutations
Combinations
Example: How many 5-card poker hands will have 3 aces and
2 kings?
The solution involves both the multiplication principle and
combinations.
O1 : Choose 3 aces out of 4 N1 : C4,3
O2 : Choose 2 kings out of 4 N2 : C4,2
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Jason Aubrey Math 1300 Finite Mathematics
46. Factorials
Permutations
Combinations
Example: How many 5-card poker hands will have 3 aces and
2 kings?
The solution involves both the multiplication principle and
combinations.
O1 : Choose 3 aces out of 4 N1 : C4,3
O2 : Choose 2 kings out of 4 N2 : C4,2
Using the multiplication principle, we have:
number of hands = C4,3 C4,2
4! 4!
=
3!(4 − 3)! 2!(4 − 2)!
= 4 · 6 = 24
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Jason Aubrey Math 1300 Finite Mathematics
47. Factorials
Permutations
Combinations
Example: A computer store receives a shipment of 24 laser
printers, including 5 that are defective. Three of these printers
are selected to be displayed in the store.
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Jason Aubrey Math 1300 Finite Mathematics
48. Factorials
Permutations
Combinations
Example: A computer store receives a shipment of 24 laser
printers, including 5 that are defective. Three of these printers
are selected to be displayed in the store.
(a) How many selections can be made?
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Jason Aubrey Math 1300 Finite Mathematics
49. Factorials
Permutations
Combinations
Example: A computer store receives a shipment of 24 laser
printers, including 5 that are defective. Three of these printers
are selected to be displayed in the store.
(a) How many selections can be made?
P(24, 3)
C(24, 3) =
3!
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Jason Aubrey Math 1300 Finite Mathematics
50. Factorials
Permutations
Combinations
Example: A computer store receives a shipment of 24 laser
printers, including 5 that are defective. Three of these printers
are selected to be displayed in the store.
(a) How many selections can be made?
P(24, 3)
C(24, 3) =
3!
24 · 23 · 22
=
3·2·1
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Jason Aubrey Math 1300 Finite Mathematics
51. Factorials
Permutations
Combinations
Example: A computer store receives a shipment of 24 laser
printers, including 5 that are defective. Three of these printers
are selected to be displayed in the store.
(a) How many selections can be made?
P(24, 3)
C(24, 3) =
3!
24 · 23 · 22
=
3·2·1
= 2024
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Jason Aubrey Math 1300 Finite Mathematics
52. Factorials
Permutations
Combinations
(b) How many of these selections will contain no defective
printers?
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Jason Aubrey Math 1300 Finite Mathematics
53. Factorials
Permutations
Combinations
(b) How many of these selections will contain no defective
printers?
P(19, 3)
C(19, 3) =
3!
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Jason Aubrey Math 1300 Finite Mathematics
54. Factorials
Permutations
Combinations
(b) How many of these selections will contain no defective
printers?
P(19, 3)
C(19, 3) =
3!
19 · 18 · 17
=
3·2·1
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Jason Aubrey Math 1300 Finite Mathematics
55. Factorials
Permutations
Combinations
(b) How many of these selections will contain no defective
printers?
P(19, 3)
C(19, 3) =
3!
19 · 18 · 17
=
3·2·1
= 969
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Jason Aubrey Math 1300 Finite Mathematics
56. Factorials
Permutations
Combinations
Example: Suppose a certain bank has 4 branches in
Columbia, 12 branches in St. Louis, and 10 branches in
Kansas City. The bank must close 6 of these branches.
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Jason Aubrey Math 1300 Finite Mathematics
57. Factorials
Permutations
Combinations
Example: Suppose a certain bank has 4 branches in
Columbia, 12 branches in St. Louis, and 10 branches in
Kansas City. The bank must close 6 of these branches.
(a) In how many ways can this be done?
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Jason Aubrey Math 1300 Finite Mathematics
58. Factorials
Permutations
Combinations
Example: Suppose a certain bank has 4 branches in
Columbia, 12 branches in St. Louis, and 10 branches in
Kansas City. The bank must close 6 of these branches.
(a) In how many ways can this be done?
C(26, 6) = 230, 230
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Jason Aubrey Math 1300 Finite Mathematics
59. Factorials
Permutations
Combinations
Example: Suppose a certain bank has 4 branches in
Columbia, 12 branches in St. Louis, and 10 branches in
Kansas City. The bank must close 6 of these branches.
(a) In how many ways can this be done?
C(26, 6) = 230, 230
(b) The bank decides to close 2 branches in Columbia, 3
branches in St. Louis, and 1 branch in Kansas City. In how
many ways can this be done?
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Jason Aubrey Math 1300 Finite Mathematics
60. Factorials
Permutations
Combinations
Example: Suppose a certain bank has 4 branches in
Columbia, 12 branches in St. Louis, and 10 branches in
Kansas City. The bank must close 6 of these branches.
(a) In how many ways can this be done?
C(26, 6) = 230, 230
(b) The bank decides to close 2 branches in Columbia, 3
branches in St. Louis, and 1 branch in Kansas City. In how
many ways can this be done?
O1 : pick Columbia branches N1 : C(4, 2)
O2 : pick STL branches N2 : C(12, 3)
O3 : pick KC branches N3 : C(10, 1)
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Jason Aubrey Math 1300 Finite Mathematics
62. Factorials
Permutations
Combinations
O1 : pick Columbia branches N1 : C(4, 2)
O2 : pick STL branches N2 : C(12, 3)
O3 : pick KC branches N3 : C(10, 1)
By the multiplication principle
Number of ways = C(4, 2)C(12, 3)C(10, 1)
= 6 · 220 · 10
= 13, 200
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Jason Aubrey Math 1300 Finite Mathematics
63. Factorials
Permutations
Combinations
Example (FS09 Final Exam) During the semester, a professor
wrote various problems to possibly include on the final exam in
a math course. The course covered 4 chapters of a textbook.
She wrote 2 problems from Chapter 1, 3 problems from
Chapter 2, 5 problems from Chapter 3, and 4 problems from
Chapter 4. She decides to put two questions from each chapter
on the final exam.
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Jason Aubrey Math 1300 Finite Mathematics
64. Factorials
Permutations
Combinations
Example (FS09 Final Exam) During the semester, a professor
wrote various problems to possibly include on the final exam in
a math course. The course covered 4 chapters of a textbook.
She wrote 2 problems from Chapter 1, 3 problems from
Chapter 2, 5 problems from Chapter 3, and 4 problems from
Chapter 4. She decides to put two questions from each chapter
on the final exam.
(a) In how many different ways can she choose two questions
from each chapter?
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Jason Aubrey Math 1300 Finite Mathematics
67. Factorials
Permutations
Combinations
(b) Suppose she writes the final by (1) choosing a set of 8
questions as above, and then (2) randomly ordering those 8
questions. Accounting for both operations, how many distinct
final exams are possible?
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Jason Aubrey Math 1300 Finite Mathematics
68. Factorials
Permutations
Combinations
(b) Suppose she writes the final by (1) choosing a set of 8
questions as above, and then (2) randomly ordering those 8
questions. Accounting for both operations, how many distinct
final exams are possible?
O1 choose questions as in (a) N1 = 180
O2 put these questions in random order
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Jason Aubrey Math 1300 Finite Mathematics
69. Factorials
Permutations
Combinations
(b) Suppose she writes the final by (1) choosing a set of 8
questions as above, and then (2) randomly ordering those 8
questions. Accounting for both operations, how many distinct
final exams are possible?
O1 choose questions as in (a) N1 = 180
O2 put these questions in random order N2 = P(8, 8)
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Jason Aubrey Math 1300 Finite Mathematics
70. Factorials
Permutations
Combinations
(b) Suppose she writes the final by (1) choosing a set of 8
questions as above, and then (2) randomly ordering those 8
questions. Accounting for both operations, how many distinct
final exams are possible?
O1 choose questions as in (a) N1 = 180
O2 put these questions in random order N2 = P(8, 8)
So, by the multiplication principle:
180 · P(8, 8) = 180 · 8! = 7, 257, 600
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Jason Aubrey Math 1300 Finite Mathematics
71. Factorials
Permutations
Combinations
(c) Suppose the questions must be in order by chapter. For
example, the Chapter 1 questions must come before the
Chapter 2 questions, the Chapter 2 questions must come
before the Chapter 3 questions, etc. In this case, how many
different final exams are possible?
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Jason Aubrey Math 1300 Finite Mathematics
72. Factorials
Permutations
Combinations
(c) Suppose the questions must be in order by chapter. For
example, the Chapter 1 questions must come before the
Chapter 2 questions, the Chapter 2 questions must come
before the Chapter 3 questions, etc. In this case, how many
different final exams are possible?
O1 choose Ch1 questions N1 = P(2, 2)
O2 choose Ch2 questions N2 = P(3, 2)
O3 choose Ch3 questions N3 = P(5, 2)
O4 choose Ch4 questions N4 = P(4, 2)
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Jason Aubrey Math 1300 Finite Mathematics
73. Factorials
Permutations
Combinations
(c) Suppose the questions must be in order by chapter. For
example, the Chapter 1 questions must come before the
Chapter 2 questions, the Chapter 2 questions must come
before the Chapter 3 questions, etc. In this case, how many
different final exams are possible?
O1 choose Ch1 questions N1 = P(2, 2)
O2 choose Ch2 questions N2 = P(3, 2)
O3 choose Ch3 questions N3 = P(5, 2)
O4 choose Ch4 questions N4 = P(4, 2)
So, by the multiplication principle:
P(2, 2)P(3, 2)P(5, 2)P(4, 2) = 2, 880
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Jason Aubrey Math 1300 Finite Mathematics
