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Basic Permutations 3Basic Permutations 3
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Question 1 of 5
1. Question
A diner offers `4` different meals. How many ways can we have `2` different meals at a time?- (12)
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 meal:We can choose from any of the `4` meals`=``4`Second meal:One meal has already been chosen. Hence we are left with `3` meals`=``3`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `4``times``3` `=` `12` There are `12` ways to have `2` different meals.`12`Method TwoWe can only have `2` different meals `(r)` from a menu of `4` meals `(n)``r=2``n=4`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{4}P_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{2})!}$$ Substitute the values of `n` and `r` `=` $$\frac{4!}{2!}$$ `=` $$\frac{4\cdot3\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$4\cdot3$$ Cancel like terms `=` `12` There are `12` ways to have `2` different meals.`12` -
Question 2 of 5
2. Question
How many ways can we select `2` flags from a set of `7` different flags?- (42)
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 stageFirst flag:We can choose from any of the `7` flags`=``7`Second flag:One flag has already been chosen. Hence we are left with `6` flags`=``6`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `7``times``6` `=` `42` There are `42` ways to select `2` flags from a set of `7` flags.`42`Method TwoWe need to select `2` flags `(r)` from `7` different flags `(n)``r=2``n=7`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{7}P_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{2})!}$$ Substitute the values of `n` and `r` `=` $$\frac{7!}{5!}$$ `=` $$\frac{7\cdot6\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$7\cdot6$$ Cancel like terms `=` `42` There are `42` ways to select `2` flags from a set of `7` flags.`42` -
Question 3 of 5
3. Question
How many ways can we select `4` flags from a set of `7` different flags?- (840)
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 flag:We can choose from any of the `7` flags`=``7`Second flag:One flag has already been chosen. Hence we are left with `6` flags`=``6`Third flag:Two flags have already been chosen. Hence we are left with `5` flags`=``5`Fourth flag:Three flags have already been chosen. Hence we are left with `4` flags`=``4`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` `=` `840` There are `840` ways to select `4` flags from a set of `7` flags.`840`We need to select `4` flags `(r)` from `7` different flags `(n)``r=4``n=7`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{7}P_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{4})!}$$ Substitute the values of `n` and `r` `=` $$\frac{7!}{3!}$$ `=` $$\frac{7\cdot6\cdot5\cdot4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$7\cdot6\cdot5\cdot4$$ Cancel like terms `=` `840` There are `840` ways to select `4` flags from a set of `7` flags.`840` -
Question 4 of 5
4. Question
Find the value of `r`$$_8 P_r=336$$- `r=` (3)
Hint
Help VideoCorrect
Great Work!
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, identify the values in the formula.$$_\color{purple}{8}P_{\color{green}{r}}$$ `=` `336` `r` `=` `r` `n` `=` `8` Substitute the values into the permutation formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ $$_\color{purple}{8}P_{\color{green}{r}}$$ `=` `336` Substitute the values of `n` and `r` $$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{r})!}$$ `=` `336` Permutation Formula `336(8-r)!` `=` $$8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$$ Cross multiply `336(8-r)!``divide336` `=` `40320``divide336` Divide both sides by `336` `(8-r)!` `=` `120` `(8-r)!` `=` `5!` `120=5xx4xx3xx2xx1` `8-r` `=` `5` Cancel the factorial from both sides `8-r` `-8` `=` `5` `-8` Subtract `5` from both sides `-r``times(-1)` `=` `-3``times(-1)` Multiply both sides by `-1` `r` `=` `3` `r=3` -
Question 5 of 5
5. Question
Find the value of `r`$$_7 P_r=840$$- (4)
Hint
Help VideoCorrect
Exceptional!
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, identify the values in the formula.$$_\color{purple}{7}P_{\color{green}{r}}$$ `=` `840` `r` `=` `r` `n` `=` `7` Substitute the values into the permutation formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ $$_\color{purple}{7}P_{\color{green}{r}}$$ `=` `840` Substitute the values of `n` and `r` $$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{r})!}$$ `=` `840` Permutation Formula `840(7-r)!` `=` $$7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$$ Cross multiply `840(7-r)!``divide840` `=` `5040``divide840` Divide both sides by `840` `(7-r)!` `=` `6` `(7-r)!` `=` `3!` `6=3xx2xx1` `7-r` `=` `3` Cancel the factorial from both sides `7-r` `-7` `=` `3` `-7` Subtract `5` from both sides `-r``times(-1)` `=` `-4``times(-1)` Multiply both sides by `-1` `r` `=` `4` `r=4`
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