Permutations with Restrictions 3

Question 1 of 7

How many arrangements can be made with the letters in

$INDEPENDENCE$ if the vowels must always be together?

1. $12 400$
2. $3360$
3. $16 800$
4. $5490$

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Permutation with Repetition

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

$n$

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

$=$ count of each repeated items

Solve the permutation for the letters if the vowels are treated as a single letter and the permutation for the vowels, then multiply them.

First, treat the vowels as a single letter and list down the letters that are repeated.

$N$ $D$

$P$ $N$ $D$ $N$

$C$

$I E E E E$

$n=8$

$N$ is repeated

$3$ times

$a=3$

$D$ is repeated

$2$ times

$b=2$

Apply the formula

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

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

	$=$	$\frac{8\cdot7\cdot\color{#CC0000}{6}\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}{\color{#CC0000}{3\cdot2\cdot1\cdot2\cdot1}}$

	$=$	$87543$	Cancel like terms

	$=$	$3360$

Hence, there are

$3360$ ways to arrange

$NDPNDNC$ and

$I E E E E$

Next, solve for the permutation of the vowels.

$I$ $E E E E$

$n=5$

$E$ is repeated

$4$ times

$a=4$

Apply the formula

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

	$=$	$\frac{\color{#9a00c7}{5}!}{\color{#004ec4}{4}!}$	Substitute $n$ and $a$

	$=$	$\frac{5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{4\cdot3\cdot2\cdot1}}$

	$=$	$5$	Cancel like terms

Hence, there are

$5$ ways to arrange

$I E E E E$

Finally, multiply the two solved permutations

first permutation

$=3360$

second permutation

$=5$

$3360\times5$

$=$

$16 800$

Therefore, there are $16 800$ ways of arranging the letters

$INDEPENDENCE$ if the vowels must always be together

$16 800$

Question 2 of 7

How many arrangements can be made with the letters in

$INDEPENDENCE$ if the

$E$ ’s and the

$N$ ’s must always be grouped together?

1. $2520$
2. $4600$
3. $30 240$
4. $84$

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Permutation with Repetition

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

$n$

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

$=$ count of each repeated items

Solve three permutations: arrangement of all letters if the

$E$ ’s and

$N$ ’s are each treated as a single letter, arrangement of

$E$ ’s and arrangement of

$N$ ’s, then multiply them.

First, treat the

$E$ ’s and

$N$ ’s as a single letter each and list down the letters that are repeated.

$D$

$P$ $D$

$C$

$I$

$E E E E$

$N N N$

$n=7$

$D$ is repeated

$2$ times

$a=2$

Apply the formula

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

	$=$	$\frac{\color{#9a00c7}{7}!}{\color{#004ec4}{2}!}$	Substitute $n$ and $a$

	$=$	$\frac{7\cdot6\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$

	$=$	$76543$	Cancel like terms

	$=$	$2520$

Hence, there are

$2520$ ways to arrange

$DPDCI$ ,

$E E E E$ and

$N N N$

Next, solve for the permutation of the

$E$ ’s

$E E E E$

$n=4$

$E$ is repeated

$4$ times

$a=4$

Apply the formula

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

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

	$=$	$1$	Cancel like terms

Hence, there is only

$1$ way to arrange

$E E E E$

Then, solve for the permutation of the

$N$ ’s

$N N N$

$n=3$

$N$ is repeated

$3$ times

$a=3$

Apply the formula

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

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

	$=$	$1$	Cancel like terms

Hence, there is only

$1$ way to arrange

$N N N$

Finally, multiply the three solved permutations

first permutation

$=2520$

second permutation

$=1$

third permutation

$=1$

$2520\times1\times1$

$=$

$2520$

Therefore, there are $2520$ ways of arranging the letters

$INDEPENDENCE$ if the

$E$ ’s and

$N$ ’s must always be together

$2520$

Question 3 of 7

A license plate contains

$4$ letters. What is the probability that a license plate picked at random would have the letters

$A,A,B$ and

$C$ in any order?

1. $12/(456 976)$
2. $1/5697$
3. $32/(298 432)$
4. $2/13$

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Permutation with Repetition

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

$n$

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

$=$ count of each repeated items

Probability

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

Solve the number of ways

$A,A,B$ and

$C$ can be arranged and divide it by the total number of ways

$4$ letters can be arranged using any letters.

First, solve for the permutation of

$A,A,B$ and

$C$ .

$A$ $A$

$B$

$C$

$n=4$

$A$ is repeated

$2$ times

$a=2$

Apply the formula

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

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

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

	$=$	$4*3$	Cancel like terms

	$=$	$12$

Hence, there are $12$ ways to arrange

$A A BC$ . This is the favourable outcome

Next, count the number of letters that can be chosen for each letter in the license plate

Remember that the letters can be repeated

First to Fourth letter:

We can choose any letter from the alphabet and have it repeated. Hence, we have

$26$ choices for each letter

$=$ $26$

Use the Fundamental Counting Principle and multiply each of the options per place value.

number of ways	$=$	$m$ $times$ $n$	Fundamental Counting Principle
	$=$	$26$ $times$ $26$ $times$ $26$ $times$ $26$
	$=$	$26^4$
	$=$	$456 976$

There are $456 976$ ways of choosing four letters from the alphabet with repetition.
This is the total outcome

Finally, solve for the probability

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

	$=$	$\frac{\color{#e65021}{12}}{\color{#007DDC}{456\:976}}$

The probability that the license plate will have the letters

$A A BC$ is

$12/(456 976)$

Question 4 of 7

Find the probability of having

$I$ and

$E$ together in arranging the letters in the word

$SHIELD$

1. $1/6$
2. $1/3$
3. $6/196$
4. $32/249$

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Permutation Formula
if $(n=r)$

$_\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}!$

Probability

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

Solve the number of ways

$IE$ can be together in the word

$SHIELD$ and divide it by the total arrangements for the letters in

$SHIELD$ .

First, treat

$IE$ as a single letter. This leaves us with only $5$ letters $(r)$ to be arranged in $5$ positions $(n)$

$n=r=5$

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

$_\color{purple}{5}P_{\color{purple}{5}}$	$=$	$\color{purple}{5}!$	Substitute the value of $n$

	$=$	$5\cdot4\cdot3\cdot2\cdot1$

	$=$	$120$

There are

$120$ ways to arrange

$SHLD$ and

$IE$

Next, count all possible arrangements for

$IE$ . This means $2$ letters $(r)$ are to be arranged in $2$ positions $(n)$

$n=r=2$

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

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

	$=$	$2\cdot1$

	$=$	$2$

There are

$2$ ways to arrange

$I$ and

$E$

Multiply the two solved permutations

_$5$

$P$ _$5$

$=120$

_$2$

$P$ _$2$

$=2$

$120\cdot2$

$=$

$240$

Hence, there are $240$ ways of arranging

$SHIELD$ if

$IE$ must always be together.
This is the favourable outcome

Next, count all arrangements for the letters in

$SHIELD$ . This means we are arranging $6$ letters $(r)$ in $6$ positions $(n)$

$n=r=6$

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

$_\color{purple}{6}P_{\color{purple}{6}}$	$=$	$\color{purple}{6}!$	Substitute the value of $n$

	$=$	$6\cdot5\cdot4\cdot3\cdot2\cdot1$

	$=$	$720$

There are $720$ ways to arrange the letters in

$SHIELD$ . This is the total outcome

Finally, solve for the probability

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

	$=$	$\frac{\color{#e65021}{240}}{\color{#007DDC}{720}}$

	$=$	$\frac{1}{3}$

The probability that the letters

$IE$ will be together in arranging the letters in

$SHIELD$ is

$1/3$

Question 5 of 7

How many ways can six books be arranged on a shelf if one of them must always be kept at the end?

(120)

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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 if $(n=r)$

$_\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}!$

Since one book should be left at the end, we end up just arranging $5$ books $(r)$ in $5$ positions $(n)$

$n=r=5$

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

$_\color{purple}{5}P_{\color{purple}{5}}$	$=$	$\color{purple}{5}!$	Substitute the value of $n$

	$=$	$5\cdot4\cdot3\cdot2\cdot1$

	$=$	$120$

There are

$120$ ways to arrange six books on a shelf if one of them must be left at the end.

$120$

Question 6 of 7

How many ways can six books be arranged on a shelf if two of them must be kept together?

(240)

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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 if $(n=r)$

$_\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}!$

Solve the number of arrangements if the two books are treated as a single book and the number of arrangements for those two books, then multiply them.

First, treat the two books that must be together as a single book. This leaves us with only $5$ books $(r)$ to be arranged in $5$ positions $(n)$

$n=r=5$

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

$_\color{purple}{5}P_{\color{purple}{5}}$	$=$	$\color{purple}{5}!$	Substitute the value of $n$

	$=$	$5\cdot4\cdot3\cdot2\cdot1$

	$=$	$120$

There are

$120$ ways to arrange

$5$ books.

Next, count all possible arrangements for the two books. This means $2$ books $(r)$ are to be arranged in $2$ positions $(n)$

$n=r=2$

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

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

	$=$	$2\cdot1$

	$=$	$2$

There are

$2$ ways to arrange two books

Finally, multiply the two solved permutations

_$5$

$P$ _$5$

$=120$

_$2$

$P$ _$2$

$=2$

$120\cdot2$

$=$

$240$

Therefore, there are $240$ ways of arranging six books on a shelf is two of them must always be together

$240$

Question 7 of 7

Find the number of ways

$4$ women and

$2$ men can be arranged in a straight line, given that the

$2$ men must be on both ends

(48)

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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 if $(n=r)$

$_\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}!$

Solve the number of arrangements for the

$4$ women and the number of arrangements for the

$2$ men on both ends, then multiply them.

First, arrange $4$ women $(r)$ in $4$ places in the straight line $(n)$

$n=r=4$

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

$_\color{purple}{4}P_{\color{purple}{4}}$	$=$	$\color{purple}{4}!$	Substitute the value of $n$

	$=$	$4\cdot3\cdot2\cdot1$

	$=$	$24$

There are

$24$ ways to arrange the

$4$ women.

Next, arrange the $2$ men $(r)$ on the $2$ ends of the line $(n)$

$n=r=2$

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

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

	$=$	$2\cdot1$

	$=$	$2$

There are

$2$ ways to arrange two men

Finally, multiply the two solved permutations

_$4$

$P$ _$4$

$=24$

_$2$

$P$ _$2$

$=2$

$24\cdot2$

$=$

$48$

Therefore, there are $48$ ways of arranging

$4$ women and

$2$ men if the men are at both ends

$48$

Permutations with Restrictions 3

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Quiz summary

Information

1. Question

Hint

Well Done!

Permutation with Repetition

2. Question

Hint

Great Work!

Permutation with Repetition

3. Question

Hint

Nice Job!

Permutation with Repetition

Probability

4. Question

Hint

Keep Going!

Permutation Formula
if $(n=r)$

Probability

5. Question

Hint

Excellent!

Permutation Formula if $(n=r)$

6. Question

Hint

Fantastic!

Permutation Formula if $(n=r)$

7. Question

Hint

Great Work!

Permutation Formula if $(n=r)$

Quizzes

Permutations with Restrictions 3

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Quiz summary

Information

1. Question

Hint

Well Done!

Permutation with Repetition

2. Question

Hint

Great Work!

Permutation with Repetition

3. Question

Hint

Nice Job!

Permutation with Repetition

Probability

4. Question

Hint

Keep Going!

Permutation Formula if (n=r)(n=r)

Probability

5. Question

Hint

Excellent!

Permutation Formula if (n=r)(n=r)

6. Question

Hint

Fantastic!

Permutation Formula if (n=r)(n=r)

7. Question

Hint

Great Work!

Permutation Formula if (n=r)(n=r)

Quizzes

Permutation Formula
if $(n=r)$

Permutation Formula if $(n=r)$

Permutation Formula if $(n=r)$

Permutation Formula if $(n=r)$