Permutation Problems 2

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Question 1 of 5

How many ways can

8

$8$ paintings be arranged in a row?

(40320)

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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 position in the row

First position:

There are

8

$8$ paintings to choose from

=

$=$ $8$ $8$

Second position:

From

8

$8$ paintings, one has already been chosen. Hence, we are left with

7

$7$ choices

=

$=$ $7$ $7$

Third position:

From

8

$8$ paintings, two has already been chosen. Hence, we are left with

6

$6$ choices

=

$=$ $6$ $6$

Fourth position:

From

8

$8$ paintings, three has already been chosen. Hence, we are left with

5

$5$ choices

=

$=$ $5$ $5$

Fifth position:

From

8

$8$ paintings, four has already been chosen. Hence, we are left with

4

$4$ choices

=

$=$ $4$ $4$

Sixth position:

From

8

$8$ paintings, five has already been chosen. Hence, we are left with

3

$3$ choices

=

$=$ $3$ $3$

Seventh position:

From

8

$8$ paintings, six has already been chosen. Hence, we are left with

2

$2$ choices

=

$=$ $2$ $2$

Eighth position:

From

8

$8$ paintings, seven has already been chosen. Hence, we are left with

1

$1$ choice

=

$=$ $1$ $1$

Use the Fundamental Counting Principle and multiply the number of options for each row.

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$8$ $8$ $\times$ $times$ $7$ $7$ $\times$ $times$ $6$ $6$ $\times$ $times$ $5$ $5$ $\times$ $times$ $4$ $4$ $\times$ $times$ $3$ $3$ $\times$ $times$ $2$ $2$ $\times$ $times$ $1$ $1$
	$=$ $=$	$40320$ $40 320$

There are $40320$ $40 320$ ways to arrange

8

$8$ paintings

40320

$40 320$

Method Two

We are arranging $8$ $8$ paintings $(r)$ $(r)$ for $8$ $8$ positions in a row $(n)$ $(n)$

r = 8

$r=8$

n = 8

$n=8$

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

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

	$=$ $=$	$\frac{8!}{0!}$

	$=$ $=$	$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{1}$	$0! = 1$ $0! =1$

	$=$ $=$	$40320$ $40 320$

There are $40320$ $40 320$ ways to arrange

8

$8$ paintings

40320

$40 320$

Question 2 of 5

Out of

8

$8$ paintings, how many ways can

5

$5$ paintings be chosen to be arranged in a row?

(6720)

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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 position in the row

First position:

There are

8

$8$ paintings to choose from

=

$=$ $8$ $8$

Second position:

From

8

$8$ paintings, one has already been chosen. Hence, we are left with

7

$7$ choices

=

$=$ $7$ $7$

Third position:

From

8

$8$ paintings, two has already been chosen. Hence, we are left with

6

$6$ choices

=

$=$ $6$ $6$

Fourth position:

From

8

$8$ paintings, three has already been chosen. Hence, we are left with

5

$5$ choices

=

$=$ $5$ $5$

Fifth position:

From

8

$8$ paintings, four has already been chosen. Hence, we are left with

4

$4$ choices

=

$=$ $4$ $4$

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
	$=$ $=$	$8$ $8$ $\times$ $times$ $7$ $7$ $\times$ $times$ $6$ $6$ $\times$ $times$ $5$ $5$ $\times$ $times$ $4$ $4$
	$=$ $=$	$6720$ $6720$

There are $6720$ $6720$ ways to arrange

5

$5$ paintings out of

8

$8$

6720

$6720$

Method Two

We are arranging $5$ $5$ paintings $(r)$ $(r)$ out of $8$ $8$ paintings $(n)$ $(n)$

r = 5

$r=5$

n = 8

$n=8$

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

$_\color{purple}{8}P_{\color{green}{8}}$	$=$ $=$	$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{5})!}$	Substitute the value of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{8!}{3!}$

	$=$ $=$	$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot\color{#CC0000}{3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}}$

	$=$	$6720$	Cancel like terms and evaluate

There are $6720$ ways to arrange

$5$ paintings out of

$8$

$6720$

Question 3 of 5

In a box containing

$8$ cards marked

$1$ -

$8$ , what is the probability of drawing

$345$ in that order?

1. $1/336$
2. $1/56$
3. $1/112$
4. $6/49$

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Probability

$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$

Permutation Formula

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

First, find the total number of ways we can 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. This is the total outcome

Remember that we want to draw the cards

$345$ in that order. This means that the favourable outcome is only $1$ .

Compute for the probability.

Probability	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{336}}$

The probability of drawing the cards

$345$ in that order is

$1/336$

Question 4 of 5

In a horse race with

$13$ competing horses, what is the probability of winning a Trifecta (successfully selecting the

$1$ st,

$2$ nd and

$3$ rd places)?

1. $5/1716$
2. $1/356$
3. $2/163$
4. $1/1716$

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Probability

$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$

Permutation Formula

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

First, find the total number of ways we can select $3$ horses $(r)$ out of $13$ $(n)$

$r=3$

$n=13$

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

$_\color{purple}{13}P_{\color{green}{3}}$	$=$	$\frac{\color{purple}{13}!}{(\color{purple}{13}-\color{green}{3})!}$	Substitute the values of $n$ and $r$

	$=$	$\frac{13!}{10!}$

	$=$	$\frac{13\cdot12\cdot11\cdot\color{#CC0000}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}$

	$=$	$1716$	Cancel like terms and evaluate

There are $1716$ ways to select

$3$ horses from

$13$ competing horses in the race. This is the total outcome

Remember that we want to successfully pick the

$1$ st,

$2$ nd and

$3$ rd ranking horse. This means that the favourable outcome is only $1$ .

Compute for the probability.

Probability	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{1716}}$

The probability of a trifecta is

$1/1716$

Question 5 of 5

5. Question
How many ways can we draw $3$ cards from a box containing $8$ cards marked with $1$ - $8$ , if each card is drawn with replacement?
- (512)
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Fundamental Counting Principle

number of ways $=$ $m$ $times$ $n$

First, count the options for each draw

Remember that the cards are being drawn with replacement

First to Third draw:

There $8$ cards in the box. Since each drawn card is replaced, all three draws have $8$ options

$=$ $8$

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

number of ways $=$ $m$ $times$ $n$ Fundamental Counting Principle

$=$ $8$ $times$ $8$ $times$ $8$

$=$ $512$

There are $512$ ways of drawing $3$ cards from the box with replacement

$512$

Permutation Problems 2

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Quiz summary

Information

1. Question

Hint

Great Work!

Permutation Formula

Fundamental Counting Principle

2. Question

Hint

Well Done!

Permutation Formula

Fundamental Counting Principle

3. Question

Hint

Excellent!

Probability

Permutation Formula

4. Question

Hint

Keep Going!

Probability

Permutation Formula

5. Question

Hint

Great Work!

Fundamental Counting Principle

Quizzes

number of ways	$=$	$m$ $times$ $n$	Fundamental Counting Principle
	$=$	$8$ $times$ $8$ $times$ $8$
	$=$	$512$