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. $3628800$ $3 628 800$
2. $4200$ $4200$
3. $151200$ $151 200$
4. $302400$ $302 400$

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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.

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

$A$ is repeated

$2$ times

$a=2$

$C$ is repeated

$2$ times

$b=2$

$E$ is repeated

$3$ times

$c=3$

The total number of letters in

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

$10$ , which means $n=10$

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 $n,a,b$ and $c$

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

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

Therefore, there are $151 200$ ways to arrange

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

$151 200$

Question 5 of 5

How many arrangements can be made with the letters in

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

$IC$ must always be together?

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

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Permutation with Repetitions

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

$n$

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

$=$ count of each repeated items

First, list down the letters that are repeated.

$E N E R G E T I C$

$E$ is repeated

$3$ times

$a=3$

The total number of letters in

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

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

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 $n,a,b$ and $c$

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

	$=$	$6720$	Cancel like terms and evaluate

Remember that

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

$n=2$

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

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

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

There are

$2$ ways of arranging

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

Finally, multiply the two solved permutations

Permutation for

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

$=6720$

Permutation for $I C$

$=2$

$6720times2$

$=$

$13 440$

There are $13 440$ ways of arranging

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

$IC$ must always be together

$13 440$

Permutations with Repetitions 1

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1. Question

Hint

Correct!

Permutation Formula

Permutation with Repetitions

2. Question

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Fantastic!

Permutation Formula

Permutation with Repetitions

3. Question

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Nice Job!

Permutation with Repetitions

4. Question

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Great Work!

Permutation with Repetitions

5. Question

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Correct!

Permutation with Repetitions

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