Basic Permutations 1
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Question 1 of 5
1. Question
How many arrangements can be made with `ABC`?- (6)
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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, count the options for each stageFirst letter:We can choose `3` letters: `A,B` and `C``=``3`Second letter:One has already been chosen from `A,B` and `C`. Hence we are left with `2` choices`=``2`Second spot:Two have already been chosen from `A,B` and `C`. Hence we are left with `1` choice`=``1`Use the Fundamental Counting Principle and multiply each draws number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `3``times``2``times``1` `=` `12` There are `6` ways to arrange `ABC`$$\color{red}{A}\color{green}{B}\color{blue}{C}$$ $$\color{red}{A}\color{blue}{C}\color{green}{B}$$ $$\color{green}{B}\color{red}{A}\color{blue}{C}$$ $$\color{blue}{C}\color{green}{B}\color{red}{A}$$ $$\color{green}{B}\color{blue}{C}\color{red}{A}$$ $$\color{blue}{C}\color{red}{A}\color{green}{B}$$ `6`Method TwoWe need to arrange `3` letters `(r)` into `3` positions `(n)``r=3``n=3`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{3}P_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{3}!}{(\color{purple}{3}-\color{green}{3})!}$$ Substitute the values of `n` and `r` `=` $$\frac{3!}{0!}$$ `=` $$3\cdot2\cdot1$$ `0! =1` `=` $$6$$ There are `6` ways to arrange `ABC`$$\color{red}{A}\color{green}{B}\color{blue}{C}$$ $$\color{red}{A}\color{blue}{C}\color{green}{B}$$ $$\color{green}{B}\color{red}{A}\color{blue}{C}$$ $$\color{blue}{C}\color{green}{B}\color{red}{A}$$ $$\color{green}{B}\color{blue}{C}\color{red}{A}$$ $$\color{blue}{C}\color{red}{A}\color{green}{B}$$ `6` -
Question 2 of 5
2. Question
Evaluate$$_4 P_2$$- (12)
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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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, identify the values in the formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$_\color{purple}{4}P_{\color{green}{2}}$$ `r` `=` `2` `n` `=` `4` Substitute the values into the permutation formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ 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 `=` $$12$$ `12` -
Question 3 of 5
3. Question
Evaluate$$_8 P_1$$- (8)
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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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, identify the values in the formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$_\color{purple}{8}P_{\color{green}{1}}$$ `r` `=` `1` `n` `=` `8` Substitute the values into the permutation formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{8}P_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{1})!}$$ Substitute the values of `n` and `r` `=` $$\frac{8!}{7!}$$ `=` $$ \frac{8\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$8$$ Cancel like terms `8` -
Question 4 of 5
4. Question
Evaluate$$_7 P_7$$- (5040)
Hint
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Fantastic!
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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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, identify the values in the formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$_\color{purple}{7}P_{\color{green}{7}}$$ `r` `=` `7` `n` `=` `7` Substitute the values into the permutation formula.$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{4}P_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{7})!}$$ Substitute the values of `n` and `r` `=` $$\frac{7!}{0!}$$ `=` $$ \frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{1}$$ `0! =1` `=` $$5040$$ `5040` -
Question 5 of 5
5. Question
How many arrangements can be made with `\text(MOUSE)`?- (120)
Hint
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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 letter:We can choose from `5` letters: `M,O,U,S` and `E``=``5`Second letter:One has already been chosen from `M,O,U,S` and `E`. Hence we are left with `4` choices`=``4`Third letter:Two have already been chosen from `M,O,U,S` and `E`. Hence we are left with `3` choices`=``3`Fourth letter:Three have already been chosen from `M,O,U,S` and `E`. Hence we are left with `2` choices`=``2`Fifth letter:Four have already been chosen from `M,O,U,S` and `E`. Hence we are left with `1` choice`=``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 `\text(MOUSE)``120`Method TwoWe need to arrange `5` letters `(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 `\text(MOUSE)``120`
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