Permutation Problems 1
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Question 1 of 6
1. Question
How many sixdigit numbers can be formed using the numbers in `2 971 835` (5040)
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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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, list down all place values and count the options for eachFirst digit:We can choose number `7` digits from `2 971 835``=``7`Second digit:From `2 971 835`, one has already been chosen for the First digit. Hence we are left with `6` choices`=``6`Third digit:From `2 971 835`, two has already been chosen for the First and Second digits. Hence we are left with `5` choices`=``5`Fourth digit:From `2 971 835`, three has already been chosen for the First, Second and Third digits. Hence we are left with `4` choices`=``4`Fifth digit:From `2 971 835`, four has already been chosen for the First, Second, Third and Fourth digits. Hence we are left with `3` choices`=``3`Sixth digit:From `2 971 835`, five has already been chosen for the First, Second, Third, Fourth and Fifth digits. Hence we are left with `2` choices`=``2`Use the Fundamental Counting Principle and multiply each category’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `7``times``6``times``5``times``4``times``3``times``2` `=` `5040` `5040` sixdigit numbers can be formed using the numbers in `2 971 835`.`5040`Method TwoSolve the problem using PermutationWe need to form `6`digit numbers `(r)` from the number `2 971 835` which has `7` digits `(n)``r=6``n=7`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{7}P_{\color{green}{6}}$$ `=` $$\frac{\color{purple}{7}!}{(\color{purple}{7}\color{green}{6})!}$$ Substitute the values of `n` and `r` `=` $$\frac{7!}{1!}$$ `=` $$\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot\color{#CC0000}{1}}{\color{#CC0000}{1}}$$ `=` $$5040$$ Cancel like terms and evaluate `5040` sixdigit numbers can be formed using the numbers in `2 971 835`.`5040` 
Question 2 of 6
2. Question
How many ways can `3` cards be drawn from a box containing `8` cards marked with numbers `1``8`? (336)
Hint
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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, list down all draws and count the options for eachFirst draw:The box contains `8` cards`=``8`Second draw:One card has already been drawn from the box. Hence we are left with `7` cards`=``7`Third draw:Two cards has already been drawn from the box. Hence we are left with `6` cards`=``6`Use the Fundamental Counting Principle and multiply each draw’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `8``times``7``times``6` `=` `336` There are `336` ways to draw `3` cards from a box containing `8` cards`336`Method TwoSolve the problem using PermutationWe need to draw `3` cards `(r)` from a box containing `8` cards `(n)``r=3``n=8`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{8}P_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{8}!}{(\color{purple}{8}\color{green}{3})!}$$ Substitute the values of `n` and `r` `=` $$\frac{8!}{5!}$$ `=` $$\frac{8\cdot7\cdot6\cdot\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1}}$$ `=` $$336$$ Cancel like terms and evaluate There are `336` ways to draw `3` cards from a box containing `8` cards`336` 
Question 3 of 6
3. Question
How many ways can a Captain and Vice Captain be chosen from `13` soccer players? (156)
Hint
Help VideoCorrect
Excellent!
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 stageChoosing the captain:There are `13` soccer players`=``13`Choosing a Vice Captain:One card has already been chosen to be the Captain. Hence we are left with `12` soccer players`=``12`Use the Fundamental Counting Principle and multiply each draw’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `13``times``12` `=` `156` There are `156` ways to choose the Captain and Vice Captain out of `13` soccer players`156`Method TwoSolve the problem using PermutationWe need to choose `2` players to be the Captain and Vice Captain `(r)` from `13` soccer players`(n)``r=2``n=13`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{13}P_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{13}!}{(\color{purple}{13}\color{green}{2})!}$$ Substitute the values of `n` and `r` `=` $$\frac{13!}{11!}$$ `=` $$\frac{13\cdot12\cdot\color{#CC0000}{11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}$$ `=` $$156$$ Cancel like terms and evaluate There are `156` ways to choose the Captain and Vice Captain out of `13` soccer players`156` 
Question 4 of 6
4. Question
How many ways can a complete soccer team of `11` players be formed from `13` soccer players?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 player:There are `13` soccer players to choose from`=``13`Second player:From `13` players, one has already been chosen. Hence, we are left with `12` choices`=``12`Third player:From `13` players, two has already been chosen. Hence, we are left with `11` choices`=``11`Fourth player:From `13` players, three has already been chosen. Hence, we are left with `10` choices`=``10`Fifth player:From `13` players, four has already been chosen. Hence, we are left with `9` choices`=``9`Sixth player:From `13` players, five has already been chosen. Hence, we are left with `8` choices`=``8`Seventh player:From `13` players, six has already been chosen. Hence, we are left with `7` choices`=``7`Eighth player:From `13` players, seven has already been chosen. Hence, we are left with `6` choices`=``6`Ninth player:From `13` players, eight has already been chosen. Hence, we are left with `5` choices`=``5`Tenth player:From `13` players, nine has already been chosen. Hence, we are left with `4` choices`=``4`Eleventh player:From `13` players, ten has already been chosen. Hence, we are left with `3` choices`=``3`Use the Fundamental Counting Principle and multiply each draw’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `13``times``12``times``11``times``10``times``9``times`
`8``times``7``times``6``times``5``times``4``times``3``=` `3 113 510 400` There are `3 113 510 400` ways to form a soccer team out of `13` soccer players`3 113 510 400`Method TwoSolve the problem using PermutationWe need to choose `11` players `(r)` from `13` soccer players`(n)``r=11``n=13`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{13}P_{\color{green}{11}}$$ `=` $$\frac{\color{purple}{13}!}{(\color{purple}{13}\color{green}{11})!}$$ Substitute the values of `n` and `r` `=` $$\frac{13!}{2!}$$ `=` $$\frac{13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$$ `=` `3 113 510 400` Cancel like terms and evaluate There are `3 113 510 400` ways to form a soccer team out of `13` soccer players`3 113 510 400` 
Question 5 of 6
5. Question
Out of `12` Formula `1` cars, how many ways can the first, second and third places be selected?Hint
Help VideoCorrect
Keep Going!
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 place:There are `12` cars to choose from`=``12`Second place:From `12` cars, one has already been chosen. Hence, we are left with `11` choices`=``11`Third place:From `12` cars, two has already been chosen. Hence, we are left with `10` choices`=``10`Use the Fundamental Counting Principle and multiply each draw’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `12``times``11``times``10` `=` `1320` There are `1320` ways to choose the first, second and third place out of `12` cars`1320`Method TwoWe are selecting `3` cars `(r)` out of `12` total cars `(n)``r=3``n=12`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{12}P_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{12}!}{(\color{purple}{12}\color{green}{3})!}$$ Substitute the value of `n` and `r` `=` $$\frac{12!}{9!}$$ `=` $$\frac{12\cdot11\cdot10\cdot\color{#CC0000}{9}\cdot\color{#CC0000}{8}\cdot\color{#CC0000}{7}\cdot\color{#CC0000}{6}\cdot\color{#CC0000}{5}\cdot\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}{\color{#CC0000}{9}\cdot\color{#CC0000}{8}\cdot\color{#CC0000}{7}\cdot\color{#CC0000}{6}\cdot\color{#CC0000}{5}\cdot\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}$$ `=` $$1320$$ Cancel like terms There are `1320` ways to select the first, second and third places out of `12` cars`1320` 
Question 6 of 6
6. Question
How many ways can `12` Formula `1` cars be arranged in ranking?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 rankingFirst place:There are `12` cars to choose from`=``12`Second place:From `12` cars, one has already been chosen. Hence, we are left with `11` choices`=``11`Third place:From `12` cars, two has already been chosen. Hence, we are left with `10` choices`=``10`Fourth place:From `12` cars, three has already been chosen. Hence, we are left with `9` choices`=``9`Fifth place:From `12` cars, four has already been chosen. Hence, we are left with `8` choices`=``8`Sixth place:From `12` cars, five has already been chosen. Hence, we are left with `7` choices`=``7`Seventh place:From `12` cars, six has already been chosen. Hence, we are left with `6` choices`=``6`Eighth place:From `12` cars, seven has already been chosen. Hence, we are left with `5` choices`=``5`Ninth place:From `12` cars, eight has already been chosen. Hence, we are left with `4` choices`=``4`Tenth place:From `12` cars, nine has already been chosen. Hence, we are left with `3` choices`=``3`Eleventh place:From `12` cars, ten has already been chosen. Hence, we are left with `2` choices`=``2`Twelfth place:From `12` cars, eleven has already been chosen. Hence, we are left with `1` choice`=``1`Use the Fundamental Counting Principle and multiply the options for each rankingnumber of ways `=` `m``times``n` Fundamental Counting Principle `=` `12``times``11``times``10``times``9``times``8``times``7`
`times``6``times``5``times``4``times``3``times``2``times``1``=` `479 001 600` There are `479 001 600` ways to rank `12` cars`479 001 600`Method TwoWe are ranking `12` cars `(r)` out of `12` total cars `(n)``r=12``n=12`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{12}P_{\color{green}{12}}$$ `=` $$\frac{\color{purple}{12}!}{(\color{purple}{12}\color{green}{12})!}$$ Substitute the value of `n` and `r` `=` $$\frac{12!}{0!}$$ `=` $$\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{1}$$ `0! =1` `=` `479 001 600` There are `479 001 600` ways to rank `12` cars`479 001 600`
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