08 Permutations and Combinations Problem Set
PERMUTATIONS AND COMBINATIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: There are 7 snowboarders competing in a competition. Assuming that there are no ties, the number of ways that the gold, silver, and bronze medals could be awarded to this group is most close to: A. 120 B. 140 C. 180 D. 210
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SOLUTION 1: The first step in this problem is recognizing that we are dealing with a permutation, as order must be considered. The FORMULA for PERMUTATIONS can be referenced under the SUBJECT of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition So defining π and π, we find that: β’ π = 7, which is the number of snowboarders β’ π = 3, which represents the number of medals
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Plugging these values into the equation:
π(7,3) =
7! 7! = = 210 (7 β 3)! 4!
There are 210 ways that the gold, silver, and bronze medals can be awarded to the 7 snowboarders.
The correct answer choice is D. πππ PROBLEM 2: Six students show up to take a fluid mechanics exam with a desk for each of them. If there are a total of 6 desks, the number ways these students can be placed at these desks is closest to: A. 720 B. 840 C. 960 D. 1020
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SOLUTION 2: The first step in this problem is recognizing that we are dealing with a permutation, as order must be considered. The formula for PERMUTATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition So defining π and π, we find that: β’ π = 6, which is the number of students β’ π = 6, which represents the number of desks Plugging these values into the equation:
π(7,3) =
6! 6! = = 720 (6 β 6)! 0! Made with
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There are 210 ways that the gold, silver, and bronze medals can be awarded to the 7 snowboarders.
Therefore, the correct answer choice is A. πππ
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PROBLEM 3: Students at the University of Florida are given a student ID with a six-digit numeral from the numbers 0 to 5 for each digit, but each numeral can only be used one. The number of unique students IDS that can be given out is most close to: A. 720 B. 840 C. 960 D. 1020
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SOLUTION 3: The first step in this problem is recognizing that we are dealing with a permutation, so order must be considered. The formula for PERMUTATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition So defining π and π, we find that: β’ π = 6, which is the number of digits on a student ID β’ π = 6, which represents the number of digits each number could be
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Plugging these values into the equation:
π(7,3) =
6! 6! = = 720 (6 β 6)! 0!
There are 720 unique IDs that can be given out.
Therefore, the correct answer choice is A. πππ
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PROBLEM 4: Given 20 objects that are placed in groups of 5 at a time, the number of possible combinations is most close to: A. 11,937 B. 12,381 C. 14,280 D. 15,504
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SOLUTION 4: As order does not matter, we know we are working with a combination problem. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π = 20 β’ π=5 Plugging these values into the equation for a combination:
πΆ(π, π) =
20! = 15,504 [5! (20 β 5)!]
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Therefore, there are 15,504 ways to arrange 20 items taken 5 at a time when order does not matter.
Therefore, the correct answer choice is D. ππ, πππ
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PROBLEM 5: The local home owners association is looking to create a committee of 7 members. If there is a pool of 17 individuals to choose from, the number of ways to select a committee of 7 members is most close to: A. 18,368 B. 19,448 C. 19,212 D. 21,347
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SOLUTION 5: As order does not matter, we know we are working with a combination problem. Committees are always a combination unless the problem states that there is a set hierarchy of the members within the committee (ie, president, vice-president, etc). If a committee is ordered in a hierarchy with a defined order to follow, then it would be considered a permutation. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π = 17 β’ π=7
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Plugging these values into the equation for a combination:
πΆ(π, π) =
17! = 19,448 [7! (17 β 7)!]
Therefore, there are 19,448 ways to form the committee.
Therefore, the correct answer choice is B. ππ, πππ
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PROBLEM 6: Given a standard deck of 52 cards, the total number of 5 card hands that can be drawn from the desk is most close to: A. 2,598,960 B. 2,847,280 C. 3,280,423 D. 3,840,232
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SOLUTION 6: When a deck of cards is dealt, order does not matter, and since order does not matter, we are dealing with a combination problem. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π = 52 β’ π=5 Plugging these values into the equation for a combination:
πΆ(π, π) =
52! = 2,598,960 [5! (52 β 5)!]
Therefore, there are 2,598,960 ways five cards can be drawn from a deck of cards.
Therefore, the correct answer choice is A. π, πππ, πππ Made with
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PROBLEM 7: The number of different arrangements that a total of 7 people can be re-arranged in with a specific order is most close to: A. 4,208 B. 5,040 C. 6,280 D. 9,280
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SOLUTION 7: The first step in this problem is recognizing that we are dealing with a permutation, so order must be considered. The formula for PERMUTATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition
π(7,7) =
π! (π β π)!
π(7,7) =
7! = 5,040 (7 β 7)!
Therefore, the correct answer choice is B. π, πππ
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PROBLEM 8: Given that the limit a Chevy Tahoe is 6 people for the vehicle, the number of arrangements for 8 different people to sit in the vehicle is most close to: A. B. C. D.
7 14 28 35
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SOLUTION 8: The answer does not depend on the seating arrangement. If it did, it would be a permutation. Hence, we can use the combination relationship. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π=8 β’ π=6 Plugging these values into the equation for a combination:
πΆ(8, 6) =
8! = 28 (8 β 6)! 6!
Therefore, the correct answer choice is C. ππ
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PROBLEM 9: Which of the following statements best describes how one would find the number of ways 10 π‘βππππ could be placed in two groups if order does not matter? A. πΉπππ‘πππππ B. πππππ’π‘ππ‘πππ C. πΆπππππππ‘πππ D. πΉπππ‘ππππ§ππ‘πππ
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SOLUTION 9: The topic of PERMUTATIONS AND COMBINATIONS can be referenced under the subject of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A combination is used to find the number of ways groups of non-ordered objects can be chosen.
Therefore, the correct answer choice is C. πͺππππππππππ
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PROBLEM 1: There are 7 snowboarders competing in a competition. Assuming that there are no ties, the number of ways that the gold, silver, and bronze medals could be awarded to this group is most close to: A. 120 B. 140 C. 180 D. 210
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SOLUTION 1: The first step in this problem is recognizing that we are dealing with a permutation, as order must be considered. The FORMULA for PERMUTATIONS can be referenced under the SUBJECT of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition So defining π and π, we find that: β’ π = 7, which is the number of snowboarders β’ π = 3, which represents the number of medals
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Plugging these values into the equation:
π(7,3) =
7! 7! = = 210 (7 β 3)! 4!
There are 210 ways that the gold, silver, and bronze medals can be awarded to the 7 snowboarders.
The correct answer choice is D. πππ PROBLEM 2: Six students show up to take a fluid mechanics exam with a desk for each of them. If there are a total of 6 desks, the number ways these students can be placed at these desks is closest to: A. 720 B. 840 C. 960 D. 1020
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SOLUTION 2: The first step in this problem is recognizing that we are dealing with a permutation, as order must be considered. The formula for PERMUTATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition So defining π and π, we find that: β’ π = 6, which is the number of students β’ π = 6, which represents the number of desks Plugging these values into the equation:
π(7,3) =
6! 6! = = 720 (6 β 6)! 0! Made with
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There are 210 ways that the gold, silver, and bronze medals can be awarded to the 7 snowboarders.
Therefore, the correct answer choice is A. πππ
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PROBLEM 3: Students at the University of Florida are given a student ID with a six-digit numeral from the numbers 0 to 5 for each digit, but each numeral can only be used one. The number of unique students IDS that can be given out is most close to: A. 720 B. 840 C. 960 D. 1020
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SOLUTION 3: The first step in this problem is recognizing that we are dealing with a permutation, so order must be considered. The formula for PERMUTATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition So defining π and π, we find that: β’ π = 6, which is the number of digits on a student ID β’ π = 6, which represents the number of digits each number could be
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Plugging these values into the equation:
π(7,3) =
6! 6! = = 720 (6 β 6)! 0!
There are 720 unique IDs that can be given out.
Therefore, the correct answer choice is A. πππ
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PROBLEM 4: Given 20 objects that are placed in groups of 5 at a time, the number of possible combinations is most close to: A. 11,937 B. 12,381 C. 14,280 D. 15,504
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SOLUTION 4: As order does not matter, we know we are working with a combination problem. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π = 20 β’ π=5 Plugging these values into the equation for a combination:
πΆ(π, π) =
20! = 15,504 [5! (20 β 5)!]
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Therefore, there are 15,504 ways to arrange 20 items taken 5 at a time when order does not matter.
Therefore, the correct answer choice is D. ππ, πππ
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PROBLEM 5: The local home owners association is looking to create a committee of 7 members. If there is a pool of 17 individuals to choose from, the number of ways to select a committee of 7 members is most close to: A. 18,368 B. 19,448 C. 19,212 D. 21,347
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SOLUTION 5: As order does not matter, we know we are working with a combination problem. Committees are always a combination unless the problem states that there is a set hierarchy of the members within the committee (ie, president, vice-president, etc). If a committee is ordered in a hierarchy with a defined order to follow, then it would be considered a permutation. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π = 17 β’ π=7
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Plugging these values into the equation for a combination:
πΆ(π, π) =
17! = 19,448 [7! (17 β 7)!]
Therefore, there are 19,448 ways to form the committee.
Therefore, the correct answer choice is B. ππ, πππ
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PROBLEM 6: Given a standard deck of 52 cards, the total number of 5 card hands that can be drawn from the desk is most close to: A. 2,598,960 B. 2,847,280 C. 3,280,423 D. 3,840,232
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SOLUTION 6: When a deck of cards is dealt, order does not matter, and since order does not matter, we are dealing with a combination problem. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π = 52 β’ π=5 Plugging these values into the equation for a combination:
πΆ(π, π) =
52! = 2,598,960 [5! (52 β 5)!]
Therefore, there are 2,598,960 ways five cards can be drawn from a deck of cards.
Therefore, the correct answer choice is A. π, πππ, πππ Made with
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PROBLEM 7: The number of different arrangements that a total of 7 people can be re-arranged in with a specific order is most close to: A. 4,208 B. 5,040 C. 6,280 D. 9,280
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SOLUTION 7: The first step in this problem is recognizing that we are dealing with a permutation, so order must be considered. The formula for PERMUTATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Formally, a permutation is the number of orders of arrangements of π objects, without repetition, selected from π distinct objects taken π at a time, given by the equation:
π(π, π) =
π! (π β π)!
Where: β’ π is the number of distinct objects β’ π is the number of objects without repetition
π(7,7) =
π! (π β π)!
π(7,7) =
7! = 5,040 (7 β 7)!
Therefore, the correct answer choice is B. π, πππ
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PROBLEM 8: Given that the limit a Chevy Tahoe is 6 people for the vehicle, the number of arrangements for 8 different people to sit in the vehicle is most close to: A. B. C. D.
7 14 28 35
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SOLUTION 8: The answer does not depend on the seating arrangement. If it did, it would be a permutation. Hence, we can use the combination relationship. The formula for COMBINATIONS can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A COMBINATION of π objects taken π at a time is an arrangement of π objects, without regard to order and without repetition, selected from π distinct objects. A combination is defined as
πΆ(π, π) =
π(π, π) π! = [π! (π β π)!] π!
From the problem, we define the following: β’ π=8 β’ π=6 Plugging these values into the equation for a combination:
πΆ(8, 6) =
8! = 28 (8 β 6)! 6!
Therefore, the correct answer choice is C. ππ
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PROBLEM 9: Which of the following statements best describes how one would find the number of ways 10 π‘βππππ could be placed in two groups if order does not matter? A. πΉπππ‘πππππ B. πππππ’π‘ππ‘πππ C. πΆπππππππ‘πππ D. πΉπππ‘ππππ§ππ‘πππ
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SOLUTION 9: The topic of PERMUTATIONS AND COMBINATIONS can be referenced under the subject of ENGINEERING PROBABILITY AND STATISTICS on page 37 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A combination is used to find the number of ways groups of non-ordered objects can be chosen.
Therefore, the correct answer choice is C. πͺππππππππππ
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