Permutations with Repetitions 1

Question 1 of 5

Find the number of ways the letters in

D A D

$DAD$ can be arranged if each letter is:

(i)

$(i)$ distinguishable

(D

$(D$ $A$ $A$ $D$ $D$

)

$)$

(i i)

$(ii)$ indistinguishable

(D A D)

$(DAD)$

$(i)$ $(i)$ (6)

$(i i)$ $(ii)$ (3)

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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)!}

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

Permutation with Repetitions

\frac{n!}{a! b! c! \dots}

$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

$n$ $n$

=

$=$ total number of items
$a, b, c$ $a,b,c$

=

$=$ count of each repeated items

(i)

$(i)$ Distinguishable letters

(D

$(D$ $A$ $A$ $D$ $D$

)

$)$

We need to arrange $3$ $3$ letters $(r)$ $(r)$ into $3$ $3$ positions $(n)$ $(n)$

r = 3

$r=3$

n = 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 $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{3!}{0!}$

	$=$ $=$	$3\cdot2\cdot1$	$0! = 1$ $0! =1$
	$=$ $=$	$6$

There are $6$ $6$ ways to arrange

D

$D$ $A$ $A$ $D$ $D$

$D$ $D$ $A$ $A$ $D$ $D$	$D$ $D$ $D$ $D$ $A$ $A$	$A$ $A$ $D$ $D$ $D$ $D$
$D$ $D$ $A$ $A$ $D$ $D$	$D$ $D$ $D$ $D$ $A$ $A$	$A$ $A$ $D$ $D$ $D$ $D$

(i i)

$(ii)$ Indistinguishable letters

(D A D)

$(DAD)$

First, list down the letters that are repeated.

D A D

$DAD$

D

$D$ is repeated

2

$2$ times

a = 2

$a=2$

The total number of letters in

D A D

$DAD$ is

3

$3$ , which means $n = 3$ $n=3$

Finally, solve for the permutation

Permutation with Repetitions	$=$ $=$	$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

	$=$ $=$	$\frac{\color{#9a00c7}{3}!}{\color{#004ec4}{2}!}$	Substitute $n$ $n$ and $a$ $a$

	$=$ $=$	$\frac{3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$

	$=$ $=$	$3$	Cancel like terms

There are $3$ $3$ ways to arrange

D A D

$DAD$

D A D

$DAD$

D D A

$DDA$

A D D

$ADD$

(i) 6

$(i) 6$

(i i) 3

$(ii) 3$

Question 2 of 5

Find the number of ways the letters in the following words can be arranged:

(i) W E A R

$(i) W E A R$

(i) W E R E

$(i) W E R E$

$(i)$ $(i)$ (24)

$(i i)$ $(ii)$ (12)

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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)!}

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

Permutation with Repetitions

\frac{n!}{a! b! c! \dots}

$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

$n$ $n$

=

$=$ total number of items
$a, b, c$ $a,b,c$

=

$=$ count of each repeated items

(i) W E A R

$(i) W E A R$

We need to arrange $4$ $4$ letters $(r)$ $(r)$ into $4$ $4$ positions $(n)$ $(n)$

r = 4

$r=4$

n = 4

$n=4$

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

$_\color{purple}{4}P_{\color{green}{4}}$	$=$ $=$	$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{4})!}$	Substitute the values of $n$ $n$ and $r$ $r$

	$=$ $=$	$\frac{4!}{0!}$

	$=$ $=$	$4\cdot3\cdot2\cdot1$	$0! = 1$ $0! =1$
	$=$ $=$	$24$

There are $24$ $24$ ways to arrange

W E A R

$W E A R$

(i i) W E R E

$(ii) WERE$

First, list down the letters that are repeated.

E

$E$ is repeated

2

$2$ times

a = 2

$a=2$

The total number of letters in

W E R E

$W E R E$ is

4

$4$ , which means $n = 4$ $n=4$

Finally, solve for the permutation

Permutation with Repetitions	$=$ $=$	$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

	$=$ $=$	$\frac{\color{#9a00c7}{4}!}{\color{#004ec4}{2}!}$	Substitute $n$ $n$ and $a$ $a$

	$=$ $=$	$\frac{4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$

	$=$ $=$	$12$	Cancel like terms and evaluate

There are $12$ $12$ ways to arrange

W E R E

$W E R E$

(i) 24

$(i) 24$

(i i) 12

$(ii) 12$

Question 3 of 5

How many arrangements can be made with the letters in

O B S E S S E D

$O B S E S S E D$ ?

(3360)

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Permutation with Repetitions

\frac{n!}{a! b! c! \dots}

$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

$n$ $n$

=

$=$ total number of items
$a, b, c$ $a,b,c$

=

$=$ count of each repeated items

First, list down the letters that are repeated.

O B S E S S E D

$O B S E S S E D$

S

$S$ is repeated

3

$3$ times

a = 3

$a=3$

E

$E$ is repeated

2

$2$ times

b = 2

$b=2$

The total number of letters in

O B S E S S E D

$O B S E S S E D$ is

8

$8$ , which means $n = 8$ $n=8$

Finally, solve for the permutation

Permutation with Repetitions	$=$ $=$	$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

	$=$ $=$	$\frac{\color{#9a00c7}{8}!}{\color{#004ec4}{3}!\color{#004ec4}{2}!}$	Substitute $n, a$ $n,a$ and $b$ $b$

	$=$ $=$	$\frac{8\cdot7\cdot\color{#CC0000}{6}\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1\cdot2\cdot1}}$

	$=$ $=$	$3360$	Cancel like terms and evaluate

Therefore, there are $3360$ $3360$ ways to arrange

O B S E S S E D

$O B S E S S E D$

3360

$3360$

Question 4 of 5

How many arrangements can be made with the letters in

A C C E L E R A T E

$A C C E L E R A T E$ ?

1. $3 628 800$ $3 628 800$
2. $4200$ $4200$
3. $302 400$ $302 400$
4. $151 200$ $151 200$

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Permutation with Repetitions

\frac{n!}{a! b! c! \dots}

$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

$n$ $n$

=

$=$ total number of items
$a, b, c$ $a,b,c$

=

$=$ count of each repeated items

First, list down the letters that are repeated.

A C C E L E R A T E

$A C C E L E R A T E$

A

$A$ is repeated

2

$2$ times

a = 2

$a=2$

C

$C$ is repeated

2

$2$ times

b = 2

$b=2$

E

$E$ is repeated

3

$3$ times

c = 3

$c=3$

The total number of letters in

A C C E L E R A T E

$A C C E L E R A T E$ is

10

$10$ , which means $n = 10$ $n=10$

Finally, solve for the permutation

Permutation with Repetitions	$=$ $=$	$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

	$=$ $=$	$\frac{\color{#9a00c7}{10}!}{\color{#004ec4}{2}!\color{#004ec4}{2}!\color{#004ec4}{3}!}$	Substitute $n, a, b$ $n,a,b$ and $c$ $c$

	$=$ $=$	$\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{2\cdot1\cdot2\cdot1\cdot3\cdot2\cdot1}}$

	$=$ $=$	$151 200$ $151 200$	Cancel like terms and evaluate

Therefore, there are $151 200$ $151 200$ ways to arrange

A C C E L E R A T E

$A C C E L E R A T E$

151 200

$151 200$

Question 5 of 5

How many arrangements can be made with the letters in

E N E R G E T I C

$E N E R G E T I C$ if

I C

$IC$ must always be together?

1. $15 120$ $15 120$
2. $41 320$ $41 320$
3. $13 440$ $13 440$
4. $6720$ $6720$

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Permutation with Repetitions

\frac{n!}{a! b! c! \dots}

$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

$n$ $n$

=

$=$ total number of items
$a, b, c$ $a,b,c$

=

$=$ count of each repeated items

First, list down the letters that are repeated.

E N E R G E T I C

$E N E R G E T I C$

E

$E$ is repeated

3

$3$ times

a = 3

$a=3$

The total number of letters in

E N E R G E T

$E N E R G E T$ $I C$ $I C$ is

8

$8$ , if $I C$ $I C$ is considered as one. This means $n = 8$ $n=8$

Solve for the permutation

Permutation with Repetitions	$=$ $=$	$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$

	$=$ $=$	$\frac{\color{#9a00c7}{8}!}{\color{#004ec4}{3}!}$	Substitute $n, a, b$ $n,a,b$ and $c$ $c$

	$=$ $=$	$\frac{8\cdot7\cdot6\cdot5\cdot4\color{#CC0000}{\cdot3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}}$

	$=$ $=$	$6720$ $6720$	Cancel like terms and evaluate

Remember that

I C

$IC$ must always be together. Find the number of ways these $2$ $2$ letters can be arranged

n = 2

$n=2$

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

$_\color{purple}{2}P_{\color{purple}{2}}$	$=$ $=$	$2!$	Substitute the value of $n$ $n$

	$=$ $=$	$2\cdot1$
	$=$ $=$	$2$

There are

2

$2$ ways of arranging

I C

$I C$ . Multiply this to the numerator of the Permutation with Repetitions

Finally, multiply the two solved permutations

Permutation for

E N E R G E T

$E N E R G E T$ $I C$ $I C$

= 6720

$=6720$

Permutation for $I C$ $I C$

= 2

$=2$

6720 \times 2

$6720times2$

=

$=$

13 440

$13 440$

There are $13 440$ $13 440$ ways of arranging

E N E R G E T I C

$E N E R G E T I C$ if

I C

$IC$ must always be together

13 440

$13 440$

Permutations with Repetitions 1

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Quiz summary

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1. Question

Hint

Correct!

Permutation Formula

Permutation with Repetitions

2. Question

Hint

Fantastic!

Permutation Formula

Permutation with Repetitions

3. Question

Hint

Nice Job!

Permutation with Repetitions

4. Question

Hint

Great Work!

Permutation with Repetitions

5. Question

Hint

Correct!

Permutation with Repetitions

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