Basic Permutations 2
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Question 1 of 5
1. Question
Peter and Erica will be checking into a hotel with `5` vacant rooms. How many ways can they be assigned to different rooms?- (20)
Hint
Help VideoCorrect
Nice Job!
Incorrect
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stagePeter’s room:Peter can be assigned to any of the `5` vacant rooms`=``5`Erica’s room:One room has already been assigned to Peter. Hence we are left with `4` vacant rooms`=``4`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `5``times``4` `=` `20` There are `20` ways to assign `2` rooms from `5` vacant rooms.`20`Method TwoWe need to find how many ways `2` rooms `(r)` can be assigned from `5` vacant rooms `(n)``r=2``n=5`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{5}P_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{2})!}$$ Substitute the values of `n` and `r` `=` $$\frac{5!}{3!}$$ `=` $$\frac{5\cdot4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$5\cdot4$$ Cancel like terms `=` $$20$$ There are `20` ways to assign `2` rooms from `5` vacant rooms.`20` -
Question 2 of 5
2. Question
How many ways can we arrange `5` tools into a tool shed?- (120)
Hint
Help VideoCorrect
Well Done!
Incorrect
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst tool:We can choose from any of the `5` tools`=``5`Second tool:One tool has already been chosen. Hence we are left with `4` tools`=``4`Third tool:Two tools have already been chosen. Hence we are left with `3` tools`=``3`Fourth tool:Three tools have already been chosen. Hence we are left with `2` tools`=``2`Fifth tool:Four tools have already been chosen. Hence we are left with `1` tool`=``1`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `5``times``4``times``3``times``2``times``1` `=` `120` There are `120` ways to arrange `5` tools.`120`Method TwoWe need to arrange `5` tools `(r)` into `5` positions `(n)``r=5``n=5`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{5}P_{\color{green}{5}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{5})!}$$ Substitute the values of `n` and `r` `=` $$\frac{5!}{0!}$$ `=` $$5\cdot4\cdot3\cdot2\cdot1$$ `0! =1` `=` $$120$$ There are `120` ways to arrange `5` tools.`120` -
Question 3 of 5
3. Question
How many ways can we assign `7` nurses as speakers at a conference?- (5040)
Hint
Help VideoCorrect
Correct!
Incorrect
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst speaker:We can choose from any of the `7` nurses`=``7`Second speaker:One nurse has already been chosen. Hence we are left with `6` nurses`=``6`Third speaker:Two nurses have already been chosen. Hence we are left with `5` nurses`=``5`Fourth speaker:Three nurses have already been chosen. Hence we are left with `4` nurses`=``4`Fifth speaker:Four nurses have already been chosen. Hence we are left with `3` nurses`=``3`Sixth speaker:Five nurses have already been chosen. Hence we are left with `2` nurses`=``2`Seventh speaker:Six nurses have already been chosen. Hence we are left with `1` nurse`=``1`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `7``times``6``times``5``times``4``times``3``times``2``times``1` `=` `5040` There are `5040` ways to assign `7` nurses as speakers at a conference.`5040`Method TwoWe need to assign `7` nurses `(r)` into `7` positions `(n)``r=7``n=7`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{7}P_{\color{green}{7}}$$ `=` $$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{7})!}$$ Substitute the values of `n` and `r` `=` $$\frac{7!}{0!}$$ `=` $$7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$$ `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)
Hint
Help VideoCorrect
Fantastic!
Incorrect
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst song:We can choose from any of the `9` songs`=``9`Second song:One song has already been chosen. Hence we are left with `8` songs`=``8`Third song:Two songs have already been chosen. Hence we are left with `7` songs`=``7`Fourth song:Three songs have already been chosen. Hence we are left with `6` songs`=``6`Fifth song:Four songs have already been chosen. Hence we are left with `5` songs`=``5`Sixth song:Five songs have already been chosen. Hence we are left with `4` songs`=``4`Seventh song:Six songs have already been chosen. Hence we are left with `3` songs`=``3`Eighth song:Seven songs have already been chosen. Hence we are left with `2` songs`=``2`Ninth song:Eight songs have already been chosen. Hence we are left with `1` song`=``1`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `9``times``8``times``7``times``6``times``5``times``4``times``3``times``2``times``1` `=` `362 880` There are `362 880` ways to arrange `9` songs on a playlist.`362 880`Method TwoWe need to arrange `9` songs `(r)` into `9` positions `(n)``r=9``n=9`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{9}P_{\color{green}{9}}$$ `=` $$\frac{\color{purple}{9}!}{(\color{purple}{9}-\color{green}{9})!}$$ Substitute the values of `n` and `r` `=` $$\frac{9!}{0!}$$ `=` $$9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$$ `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)
Hint
Help VideoCorrect
Nice Job!
Incorrect
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each stageFirst spot:We can choose from any of the `8` books`=``8`Second spot:One book has already been chosen. Hence we are left with `7` books`=``7`Third spot:Two books have already been chosen. Hence we are left with `6` books`=``6`Fourth spot:Three books have already been chosen. Hence we are left with `5` books`=``5`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `8``times``7``times``6``times``5` `=` `1680` There are `1680` ways to arrange `8` books on a shelf that can hold `4` books.`1680`Method TwoWe need to arrange `8` books `(r)` into `4` positions `(n)``r=4``n=8`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{8}P_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{4})!}$$ Substitute the values of `n` and `r` `=` $$\frac{8!}{4!}$$ `=` $$\frac{8\cdot7\cdot6\cdot5\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$8\cdot7\cdot6\cdot5$$ 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