Permutation Problems 1

Question 1 of 6

How many six-digit numbers can be formed using the numbers in

2971835

$2 971 835$

(5040)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Use the permutations formula to find the number of ways an item can be arranged $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is important in Permutations.

Permutation Formula

_{n} P_{r} = \frac{n!}{(n - r)!}

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

Fundamental Counting Principle

number of ways

=

$=$ $m$ $m$

\times

$times$ $n$ $n$

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

$7$ digits from

2971835

$2 971 835$

=

$=$ $7$ $7$

Second digit:

From

2971835

$2 971 835$ , one has already been chosen for the First digit. Hence we are left with

6

$6$ choices

=

$=$ $6$ $6$

Third digit:

From

2971835

$2 971 835$ , two has already been chosen for the First and Second digits. Hence we are left with

5

$5$ choices

=

$=$ $5$ $5$

Fourth digit:

From

2971835

$2 971 835$ , three has already been chosen for the First, Second and Third digits. Hence we are left with

4

$4$ choices

=

$=$ $4$ $4$

Fifth digit:

From

2971835

$2 971 835$ , four has already been chosen for the First, Second, Third and Fourth digits. Hence we are left with

3

$3$ choices

=

$=$ $3$ $3$

Sixth digit:

From

2971835

$2 971 835$ , five has already been chosen for the First, Second, Third, Fourth and Fifth digits. Hence we are left with

2

$2$ choices

=

$=$ $2$ $2$

Use the Fundamental Counting Principle and multiply each category’s number of options.

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$7$ $7$ $\times$ $times$ $6$ $6$ $\times$ $times$ $5$ $5$ $\times$ $times$ $4$ $4$ $\times$ $times$ $3$ $3$ $\times$ $times$ $2$ $2$
	$=$ $=$	$5040$ $5040$

$5040$ $5040$ six-digit numbers can be formed using the numbers in

2971835

$2 971 835$ .

5040

$5040$

Method Two

Solve the problem using Permutation

We need to form $6$ $6$ -digit numbers $(r)$ $(r)$ from the number

2971835

$2 971 835$ which has $7$ $7$ digits $(n)$ $(n)$

r = 6

$r=6$

n = 7

$n=7$

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

$_\color{purple}{7}P_{\color{green}{6}}$	$=$ $=$	$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{6})!}$	Substitute the values of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{7!}{1!}$

	$=$ $=$	$\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot\color{#CC0000}{1}}{\color{#CC0000}{1}}$

	$=$ $=$	$5040$	Cancel like terms and evaluate

$5040$ $5040$ six-digit numbers can be formed using the numbers in

2971835

$2 971 835$ .

5040

$5040$

Question 2 of 6

How many ways can

3

$3$ cards be drawn from a box containing

8

$8$ cards marked with numbers

1

$1$ -

8

$8$ ?

(336)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Use the permutations formula to find the number of ways an item can be arranged $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is important in Permutations.

Permutation Formula

_{n} P_{r} = \frac{n!}{(n - r)!}

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

Fundamental Counting Principle

number of ways

=

$=$ $m$ $m$

\times

$times$ $n$ $n$

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

$8$ cards

=

$=$ $8$ $8$

Second draw:

One card has already been drawn from the box. Hence we are left with

7

$7$ cards

=

$=$ $7$ $7$

Third draw:

Two cards has already been drawn from the box. Hence we are left with

6

$6$ cards

=

$=$ $6$ $6$

Use the Fundamental Counting Principle and multiply each draw’s number of options.

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$8$ $8$ $\times$ $times$ $7$ $7$ $\times$ $times$ $6$ $6$
	$=$ $=$	$336$ $336$

There are $336$ $336$ ways to draw

3

$3$ cards from a box containing

8

$8$ cards

336

$336$

Method Two

Solve the problem using Permutation

We need to draw $3$ $3$ cards $(r)$ $(r)$ from a box containing $8$ $8$ cards $(n)$ $(n)$

r = 3

$r=3$

n = 8

$n=8$

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

$_\color{purple}{8}P_{\color{green}{3}}$	$=$ $=$	$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{3})!}$	Substitute the values of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{8!}{5!}$

	$=$ $=$	$\frac{8\cdot7\cdot6\cdot\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1}}$

	$=$ $=$	$336$	Cancel like terms and evaluate

There are $336$ $336$ ways to draw

3

$3$ cards from a box containing

8

$8$ cards

336

$336$

Question 3 of 6

How many ways can a Captain and Vice Captain be chosen from

13

$13$ soccer players?

(156)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Use the permutations formula to find the number of ways an item can be arranged $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is important in Permutations.

Permutation Formula

_{n} P_{r} = \frac{n!}{(n - r)!}

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

Fundamental Counting Principle

number of ways

=

$=$ $m$ $m$

\times

$times$ $n$ $n$

Method One

Solve the problem using the Fundamental Counting Principle

First, count the options for each stage

Choosing the captain:

There are

13

$13$ soccer players

=

$=$ $13$ $13$

Choosing a Vice Captain:

One card has already been chosen to be the Captain. Hence we are left with

12

$12$ soccer players

=

$=$ $12$ $12$

Use the Fundamental Counting Principle and multiply each draw’s number of options.

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$13$ $13$ $\times$ $times$ $12$ $12$
	$=$ $=$	$156$ $156$

There are $156$ $156$ ways to choose the Captain and Vice Captain out of

13

$13$ soccer players

156

$156$

Method Two

Solve the problem using Permutation

We need to choose $2$ $2$ players to be the Captain and Vice Captain $(r)$ $(r)$ from $13$ $13$ soccer players $(n)$ $(n)$

r = 2

$r=2$

n = 13

$n=13$

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

$_\color{purple}{13}P_{\color{green}{2}}$	$=$ $=$	$\frac{\color{purple}{13}!}{(\color{purple}{13}-\color{green}{2})!}$	Substitute the values of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{13!}{11!}$

	$=$ $=$	$\frac{13\cdot12\cdot\color{#CC0000}{11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}$

	$=$ $=$	$156$	Cancel like terms and evaluate

There are $156$ $156$ ways to choose the Captain and Vice Captain out of

13

$13$ soccer players

156

$156$

Question 4 of 6

How many ways can a complete soccer team of

11

$11$ players be formed from

13

$13$ soccer players?

1. $2578$ $2578$
2. $143$ $143$
3. $3113510400$ $3 113 510 400$
4. $4823210$ $4 823 210$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Use the permutations formula to find the number of ways an item can be arranged $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is important in Permutations.

Permutation Formula

_{n} P_{r} = \frac{n!}{(n - r)!}

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

Fundamental Counting Principle

number of ways

=

$=$ $m$ $m$

\times

$times$ $n$ $n$

Method One

Solve the problem using the Fundamental Counting Principle

First, count the options for each stage

First player:

There are

13

$13$ soccer players to choose from

=

$=$ $13$ $13$

Second player:

From

13

$13$ players, one has already been chosen. Hence, we are left with

12

$12$ choices

=

$=$ $12$ $12$

Third player:

From

13

$13$ players, two has already been chosen. Hence, we are left with

11

$11$ choices

=

$=$ $11$ $11$

Fourth player:

From

13

$13$ players, three has already been chosen. Hence, we are left with

10

$10$ choices

=

$=$ $10$ $10$

Fifth player:

From

13

$13$ players, four has already been chosen. Hence, we are left with

9

$9$ choices

=

$=$ $9$ $9$

Sixth player:

From

13

$13$ players, five has already been chosen. Hence, we are left with

8

$8$ choices

=

$=$ $8$ $8$

Seventh player:

From

13

$13$ players, six has already been chosen. Hence, we are left with

7

$7$ choices

=

$=$ $7$ $7$

Eighth player:

From

13

$13$ players, seven has already been chosen. Hence, we are left with

6

$6$ choices

=

$=$ $6$ $6$

Ninth player:

From

13

$13$ players, eight has already been chosen. Hence, we are left with

5

$5$ choices

=

$=$ $5$ $5$

Tenth player:

From

13

$13$ players, nine has already been chosen. Hence, we are left with

4

$4$ choices

=

$=$ $4$ $4$

Eleventh player:

From

13

$13$ players, ten has already been chosen. Hence, we are left with

3

$3$ choices

=

$=$ $3$ $3$

Use the Fundamental Counting Principle and multiply each draw’s number of options.

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$13$ $13$ $\times$ $times$ $12$ $12$ $\times$ $times$ $11$ $11$ $\times$ $times$ $10$ $10$ $\times$ $times$ $9$ $9$ $\times$ $times$ $8$ $8$ $\times$ $times$ $7$ $7$ $\times$ $times$ $6$ $6$ $\times$ $times$ $5$ $5$ $\times$ $times$ $4$ $4$ $\times$ $times$ $3$ $3$
	$=$ $=$	$3113510400$ $3 113 510 400$

There are $3113510400$ $3 113 510 400$ ways to form a soccer team out of

13

$13$ soccer players

3113510400

$3 113 510 400$

Method Two

Solve the problem using Permutation

We need to choose $11$ $11$ players $(r)$ $(r)$ from $13$ $13$ soccer players $(n)$ $(n)$

r = 11

$r=11$

n = 13

$n=13$

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

$_\color{purple}{13}P_{\color{green}{11}}$	$=$ $=$	$\frac{\color{purple}{13}!}{(\color{purple}{13}-\color{green}{11})!}$	Substitute the values of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{13!}{2!}$

	$=$ $=$	$\frac{13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$

	$=$ $=$	$3113510400$ $3 113 510 400$	Cancel like terms and evaluate

There are $3113510400$ $3 113 510 400$ ways to form a soccer team out of

13

$13$ soccer players

3113510400

$3 113 510 400$

Question 5 of 6

Out of

12

$12$ Formula

1

$1$ cars, how many ways can the first, second and third places be selected?

1. $3113510400$ $3113510400$
2. $110$ $110$
3. $1$ $1$
4. $1320$ $1320$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Use the permutations formula to find the number of ways an item can be arranged $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is important in Permutations.

Permutation Formula

_{n} P_{r} = \frac{n!}{(n - r)!}

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

Fundamental Counting Principle

number of ways

=

$=$ $m$ $m$

\times

$times$ $n$ $n$

Method One

Solve the problem using the Fundamental Counting Principle

First, count the options for each stage

First place:

There are

12

$12$ cars to choose from

=

$=$ $12$ $12$

Second place:

From

12

$12$ cars, one has already been chosen. Hence, we are left with

11

$11$ choices

=

$=$ $11$ $11$

Third place:

From

12

$12$ cars, two has already been chosen. Hence, we are left with

10

$10$ choices

=

$=$ $10$ $10$

Use the Fundamental Counting Principle and multiply each draw’s number of options.

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$12$ $12$ $\times$ $times$ $11$ $11$ $\times$ $times$ $10$ $10$
	$=$ $=$	$1320$ $1320$

There are $1320$ $1320$ ways to choose the first, second and third place out of

12

$12$ cars

1320

$1320$

Method Two

We are selecting $3$ $3$ cars $(r)$ $(r)$ out of $12$ $12$ total cars $(n)$ $(n)$

r = 3

$r=3$

n = 12

$n=12$

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

$_\color{purple}{12}P_{\color{green}{3}}$	$=$ $=$	$\frac{\color{purple}{12}!}{(\color{purple}{12}-\color{green}{3})!}$	Substitute the value of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{12!}{9!}$

	$=$ $=$	$\frac{12\cdot11\cdot10\cdot\color{#CC0000}{9}\cdot\color{#CC0000}{8}\cdot\color{#CC0000}{7}\cdot\color{#CC0000}{6}\cdot\color{#CC0000}{5}\cdot\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}{\color{#CC0000}{9}\cdot\color{#CC0000}{8}\cdot\color{#CC0000}{7}\cdot\color{#CC0000}{6}\cdot\color{#CC0000}{5}\cdot\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}$

	$=$ $=$	$1320$	Cancel like terms

There are $1320$ $1320$ ways to select the first, second and third places out of

12

$12$ cars

1320

$1320$

Question 6 of 6

How many ways can

12

$12$ Formula

1

$1$ cars be arranged in ranking?

1. $8236$ $8236$
2. $3012845922$ $3 012 845 922$
3. $479001600$ $479 001 600$
4. $144$ $144$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Use the permutations formula to find the number of ways an item can be arranged $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is important in Permutations.

Permutation Formula

_{n} P_{r} = \frac{n!}{(n - r)!}

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

Fundamental Counting Principle

number of ways

=

$=$ $m$ $m$

\times

$times$ $n$ $n$

Method One

Solve the problem using the Fundamental Counting Principle

First, count the options for each ranking

First place:

There are

12

$12$ cars to choose from

=

$=$ $12$ $12$

Second place:

From

12

$12$ cars, one has already been chosen. Hence, we are left with

11

$11$ choices

=

$=$ $11$ $11$

Third place:

From

12

$12$ cars, two has already been chosen. Hence, we are left with

10

$10$ choices

=

$=$ $10$ $10$

Fourth place:

From

12

$12$ cars, three has already been chosen. Hence, we are left with

9

$9$ choices

=

$=$ $9$ $9$

Fifth place:

From

12

$12$ cars, four has already been chosen. Hence, we are left with

8

$8$ choices

=

$=$ $8$ $8$

Sixth place:

From

12

$12$ cars, five has already been chosen. Hence, we are left with

7

$7$ choices

=

$=$ $7$ $7$

Seventh place:

From

12

$12$ cars, six has already been chosen. Hence, we are left with

6

$6$ choices

=

$=$ $6$ $6$

Eighth place:

From

12

$12$ cars, seven has already been chosen. Hence, we are left with

5

$5$ choices

=

$=$ $5$ $5$

Ninth place:

From

12

$12$ cars, eight has already been chosen. Hence, we are left with

4

$4$ choices

=

$=$ $4$ $4$

Tenth place:

From

12

$12$ cars, nine has already been chosen. Hence, we are left with

3

$3$ choices

=

$=$ $3$ $3$

Eleventh place:

From

12

$12$ 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

$479 001 600$

Permutation Problems 1

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Well Done!

Permutation Formula

Fundamental Counting Principle

2. Question

Hint

Nice Job!

Permutation Formula

Fundamental Counting Principle

3. Question

Hint

Excellent!

Permutation Formula

Fundamental Counting Principle

4. Question

Hint

Fantastic!

Permutation Formula

Fundamental Counting Principle

5. Question

Hint

Keep Going!

Permutation Formula

Fundamental Counting Principle

6. Question

Hint

Correct!

Permutation Formula

Fundamental Counting Principle

Quizzes