Permutation Problems 1

Question 1 of 6

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

$2 971 835$

(5040)

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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 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

How many ways can

$3$ cards be drawn from a box containing

$8$ cards marked with numbers

$1$ -

$8$ ?

(336)

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Mute

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

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

$13$ soccer players?

(156)

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Mute

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

How many ways can a complete soccer team of

$11$ players be formed from

$13$ soccer players?

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

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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 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

Out of

$12$ Formula

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

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

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Mute

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

How many ways can

$12$ Formula

$1$ cars be arranged in ranking?

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

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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 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

$479 001 600$

Permutation Problems 1

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1. Question

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Well Done!

Permutation Formula

Fundamental Counting Principle

2. Question

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Nice Job!

Permutation Formula

Fundamental Counting Principle

3. Question

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Excellent!

Permutation Formula

Fundamental Counting Principle

4. Question

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Fantastic!

Permutation Formula

Fundamental Counting Principle

5. Question

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Keep Going!

Permutation Formula

Fundamental Counting Principle

6. Question

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Correct!

Permutation Formula

Fundamental Counting Principle

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