Permutation Problems 1

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Question 1 of 6

1. Question

How many six-digit numbers can be formed using the numbers in `2 971 835`

(5040)

Hint

Help Video

Correct

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 One

Solve the problem using the Fundamental Counting Principle

First, list down all place values and count the options for each

First 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` six-digit numbers can be formed using the numbers in `2 971 835`.

`5040`

Method Two

Solve the problem using Permutation

We 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` six-digit 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

Help Video

Correct

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 One

Solve the problem using the Fundamental Counting Principle

First, list down all draws and count the options for each

First 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 Two

Solve the problem using Permutation

We 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 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

Choosing 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 Two

Solve the problem using Permutation

We 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?

`3 113 510 400`
`143`
`2578`
`4 823 210`

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

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

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 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 Two

Solve the problem using Permutation

We 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?

`1320`
`1`
`110`
`3113510400`

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

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

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 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 Two

We 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?

`479 001 600`
`3 012 845 922`
`8236`
`144`

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

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

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 ranking

First 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 ranking

number 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 Two

We 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