Basic Permutations 1

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Question 1 of 5

1. Question

How many arrangements can be made with `ABC`?

Hint

Help Video

Correct

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

$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$

Fundamental Counting Principle

number of ways `=``m``times``n`

Method One

Solve the problem using the Fundamental Counting Principle

First, count the options for each stage

First 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 Two

We 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)

Hint

Help Video

Correct

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

$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$

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

$$_8 P_1$$

Hint

Help Video

Correct

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

$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$

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

Help Video

Correct

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

$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$

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

Help Video

Correct

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 One

Solve the problem using the Fundamental Counting Principle

First, count the options for each stage

First 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 Two

We 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`

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