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Basic Permutations 2Basic Permutations 2
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Question 1 of 5
1. Question
Peter and Erica will be checking into a hotel with 55 vacant rooms. How many ways can they be assigned to different rooms?- (20)
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Use the permutations formula to find the number of ways an item can be arranged (r)(r) from the total number of items (n)(n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!nPr=n!(n−r)!Fundamental Counting Principle
number of ways ==mm××nnMethod OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stagePeter’s room:Peter can be assigned to any of the 55 vacant rooms==55Erica’s room:One room has already been assigned to Peter. Hence we are left with 44 vacant rooms==44Use the Fundamental Counting Principle and multiply each draws number of options.number of ways == mm××nn Fundamental Counting Principle == 55××44 == 2020 There are 2020 ways to assign 22 rooms from 55 vacant rooms.2020Method TwoWe need to find how many ways 22 rooms (r)(r) can be assigned from 55 vacant rooms (n)(n)r=2r=2n=5n=5nPrnPr == n!(n−r)!n!(n−r)! Permutation Formula 5P25P2 == 5!(5−2)!5!(5−2)! Substitute the values of nn and rr == 5!3!5!3! == 5⋅4⋅3⋅2⋅13⋅2⋅15⋅4⋅3⋅2⋅13⋅2⋅1 == 5⋅45⋅4 Cancel like terms == 2020 There are 2020 ways to assign 22 rooms from 55 vacant rooms.2020 -
Question 2 of 5
2. Question
How many ways can we arrange 55 tools into a tool shed?- (120)
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Use the permutations formula to find the number of ways an item can be arranged (r)(r) from the total number of items (n)(n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!nPr=n!(n−r)!Fundamental Counting Principle
number of ways ==mm××nnMethod OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst tool:We can choose from any of the 55 tools==55Second tool:One tool has already been chosen. Hence we are left with 44 tools==44Third tool:Two tools have already been chosen. Hence we are left with 33 tools==33Fourth tool:Three tools have already been chosen. Hence we are left with 22 tools==22Fifth tool:Four tools have already been chosen. Hence we are left with 11 tool==11Use the Fundamental Counting Principle and multiply each draws number of options.number of ways == mm××nn Fundamental Counting Principle == 55××44××33××22××11 == 120120 There are 120120 ways to arrange 55 tools.120120Method TwoWe need to arrange 55 tools (r)(r) into 55 positions (n)(n)r=5r=5n=5n=5nPrnPr == n!(n−r)!n!(n−r)! Permutation Formula 5P55P5 == 5!(5−5)!5!(5−5)! Substitute the values of nn and rr == 5!0!5!0! == 5⋅4⋅3⋅2⋅15⋅4⋅3⋅2⋅1 0!=10!=1 == 120120 There are 120120 ways to arrange 55 tools.120120 -
Question 3 of 5
3. Question
How many ways can we assign 77 nurses as speakers at a conference?- (5040)
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Use the permutations formula to find the number of ways an item can be arranged (r)(r) from the total number of items (n)(n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!nPr=n!(n−r)!Fundamental Counting Principle
number of ways ==mm××nnMethod OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst speaker:We can choose from any of the 77 nurses==77Second speaker:One nurse has already been chosen. Hence we are left with 66 nurses==66Third speaker:Two nurses have already been chosen. Hence we are left with 55 nurses==55Fourth speaker:Three nurses have already been chosen. Hence we are left with 44 nurses==44Fifth speaker:Four nurses have already been chosen. Hence we are left with 33 nurses==33Sixth speaker:Five nurses have already been chosen. Hence we are left with 22 nurses==22Seventh speaker:Six nurses have already been chosen. Hence we are left with 11 nurse==11Use the Fundamental Counting Principle and multiply each draws number of options.number of ways == mm××nn Fundamental Counting Principle == 77××66××55××44××33××22××11 == 50405040 There are 50405040 ways to assign 77 nurses as speakers at a conference.50405040Method TwoWe need to assign 77 nurses (r)(r) into 77 positions (n)(n)r=7r=7n=7n=7nPrnPr == n!(n−r)!n!(n−r)! Permutation Formula 7P77P7 = 7!(7−7)! Substitute the values of n and r = 7!0! = 7⋅6⋅5⋅4⋅3⋅2⋅1 0!=1 = 5040 There are 5040 ways to assign 7 nurses as speakers at a conference.5040 -
Question 4 of 5
4. Question
How many ways can we arrange 9 songs on a playlist?- (362880)
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!Fundamental Counting Principle
number of ways =m×nMethod OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst song:We can choose from any of the 9 songs=9Second song:One song has already been chosen. Hence we are left with 8 songs=8Third song:Two songs have already been chosen. Hence we are left with 7 songs=7Fourth song:Three songs have already been chosen. Hence we are left with 6 songs=6Fifth song:Four songs have already been chosen. Hence we are left with 5 songs=5Sixth song:Five songs have already been chosen. Hence we are left with 4 songs=4Seventh song:Six songs have already been chosen. Hence we are left with 3 songs=3Eighth song:Seven songs have already been chosen. Hence we are left with 2 songs=2Ninth song:Eight songs have already been chosen. Hence we are left with 1 song=1Use the Fundamental Counting Principle and multiply each draws number of options.number of ways = m×n Fundamental Counting Principle = 9×8×7×6×5×4×3×2×1 = 362 880 There are 362 880 ways to arrange 9 songs on a playlist.362 880Method TwoWe need to arrange 9 songs (r) into 9 positions (n)r=9n=9nPr = n!(n−r)! Permutation Formula 9P9 = 9!(9−9)! Substitute the values of n and r = 9!0! = 9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 0!=1 = 362 880 There are 362 880 ways to arrange 9 songs on a playlist.362 880 -
Question 5 of 5
5. Question
How many ways can we arrange 8 books on a shelf that can hold 4 books at a time?- (1680)
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Chapters- Chapters
Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!Fundamental Counting Principle
number of ways =m×nMethod OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst spot:We can choose from any of the 8 books=8Second spot:One book has already been chosen. Hence we are left with 7 books=7Third spot:Two books have already been chosen. Hence we are left with 6 books=6Fourth spot:Three books have already been chosen. Hence we are left with 5 books=5Use the Fundamental Counting Principle and multiply each draws number of options.number of ways = m×n Fundamental Counting Principle = 8×7×6×5 = 1680 There are 1680 ways to arrange 8 books on a shelf that can hold 4 books.1680Method TwoWe need to arrange 8 books (r) into 4 positions (n)r=4n=8nPr = n!(n−r)! Permutation Formula 8P4 = 8!(8−4)! Substitute the values of n and r = 8!4! = 8⋅7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1 = 8⋅7⋅6⋅5 Cancel like terms = 1680 There are 1680 ways to arrange 8 books on a shelf that can hold 4 books.1680
Quizzes
- Factorial Notation
- Fundamental Counting Principle 1
- Fundamental Counting Principle 2
- Fundamental Counting Principle 3
- Combinations 1
- Combinations 2
- Combinations with Restrictions 1
- Combinations with Restrictions 2
- Combinations with Probability
- Basic Permutations 1
- Basic Permutations 2
- Basic Permutations 3
- Permutation Problems 1
- Permutation Problems 2
- Permutations with Repetitions 1
- Permutations with Repetitions 2
- Permutations with Restrictions 1
- Permutations with Restrictions 2
- Permutations with Restrictions 3
- Permutations with Restrictions 4