Permutations with Restrictions 2

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Question 1 of 7

1. Question
How many four-digit numbers can be formed WITHOUT repetition using the numbers: $1$ $1$ - $7$ $7$ ?
- 1. $720$ $720$
- 2. $490$ $490$
- 3. $360$ $360$
- 4. $5040$ $5040$
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Fundamental Counting Principle

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

Count the numbers that can be chosen for each place value

First digit:

From $1$ $1$ - $7$ $7$ , we can choose $5, 6$ $5,6$ and $7$ $7$ to make sure the number is greater than $5000$ $5000$ . Hence, we have $3$ $3$ choices

$=$ $=$ $3$ $3$

Second digit:

From $1$ $1$ - $7$ $7$ , one has already been chosen for the First digit. Hence, we are left with $6$ $6$ choices

$=$ $=$ $6$ $6$

Third digit:

From $1$ $1$ - $7$ $7$ , two has already been chosen for the First and Second digits. Hence, we are left with $5$ $5$ choices

$=$ $=$ $5$ $5$

Fourth digit:

From $1$ $1$ - $7$ $7$ , three has already been chosen for the First, Second and Third digits. Hence, we are left with $4$ $4$ choices

$=$ $=$ $4$ $4$

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

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

$=$ $=$ $3$ $3$ $\times$ $times$ $6$ $6$ $\times$ $times$ $5$ $5$ $\times$ $times$ $4$ $4$

$=$ $=$ $360$ $360$

There are $360$ $360$ ways of forming four-digit numbers greater than $5000$ $5000$ using $1$ $1$ - $7$ $7$ without repetition

$360$ $360$
Question 2 of 7

2. Question
How many four-digit numbers can be formed WITH repetition using the numbers: $1$ $1$ - $7$ $7$ ?
- 1. $5040$ $5040$
- 2. $490$ $490$
- 3. $1029$ $1029$
- 4. $3521$ $3521$
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Fundamental Counting Principle

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

Count the numbers that can be chosen for each place value

Remember that the numbers can be repeated

First digit:

From $1$ $1$ - $7$ $7$ , we can choose $5, 6$ $5,6$ and $7$ $7$ to make sure the number is greater than $5000$ $5000$ . Hence, we have $3$ $3$ choices

$=$ $=$ $3$ $3$

Second to Fourth digit:

From $1$ $1$ - $7$ $7$ , we can choose any number and have it repeated. Hence, the other three digits have $7$ $7$ choices each

$=$ $=$ $7$ $7$

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

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

$=$ $3$ $times$ $7^3$

$=$ $1029$

There are $1029$ ways of forming four-digit numbers greater than $5000$ using $1$ - $7$ with repetition

$1029$

Question 3 of 7

How many arrangements can be made with

$ABCDEFG$ if

$A$ must be placed in the middle?

1. $720$
2. $6$
3. $600$
4. $54$

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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

$A$ must be placed in the middle, we are left with only $6$ letters $(r)$ to be arranged in $6$ positions $(n)$

$_$ $_$ $_$

$A$ $_$ $_$ $_$

$n=r=6$

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

$ABCDEFG$ if

$A$ must be placed in the middle

$720$

Question 4 of 7

How many arrangements can be made with

$ABCDEFG$ if

$FG$ must always be together?

1. $600$
2. $24$
3. $720$
4. $1440$

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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 $(n=r)$

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

Multiply the number of arrangements if

$FG$ is treated as a single letter to the number or arrangements for letters

$F$ and

$G$ .

First, treat

$FG$ as a single letter. This leaves us with only $6$ other letters $(r)$ to be arranged 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

$ABCDE$ and

$FG$

Next, count all possible arrangements for

$FG$ . 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

$F$ and

$G$

Finally, multiply the two solved permutations

_$6$

$P$ _$6$

$=720$

_$2$

$P$ _$2$

$=2$

$720\cdot2$

$=$

$1440$

Therefore, there are $1440$ ways of arranging

$ABCDEFG$ if

$FG$ must always be together

$1440$

Question 5 of 7

How many arrangements can be made with

$ABCDEFG$ if

$F$ and

$G$ must not be together?

1. $720$
2. $1440$
3. $5040$
4. $3600$

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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 $(n=r)$

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

Get the difference between the total number of arrangements and the number of arrangements if

$FG$ must be together

First, count the total possible arrangements for

$ABCDEFG$

$n=r=7$

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

$_\color{purple}{7}P_{\color{purple}{7}}$	$=$	$\color{purple}{7}!$	Substitute the value of $n$

	$=$	$7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$

	$=$	$5040$

There are

$5040$ ways to arrange

$ABCDEFG$

Now, count all possible arrangements where

$FG$ must always be together.

Start with treating

$FG$ as a single letter. This leaves us with only $6$ other letters $(r)$ to be arranged 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

$ABCDE$ and

$FG$

Next, count all possible arrangements for

$FG$ . 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

$F$ and

$G$

Then, multiply _$6$

$P$ _$6$ and _$2$

$P$ _$2$

_$6$

$P$ _$6$

$=720$

_$2$

$P$ _$2$

$=2$

$720\cdot2$

$=$

$1440$

Hence, there are $1440$ ways of arranging

$ABCDEFG$ if

$FG$ must always be together

Finally, subtract the arrangement where

$FG$ must be together from the total number of arrangements

Total arrangements

$=5040$

Arrangements where

$FG$ are together

$=1440$

$5040-1440$

$=$

$3600$

Therefore, there are $3600$ ways of arranging

$ABCDEFG$ if

$F$ and

$G$ must not be together

$3600$

Question 6 of 7

How many arrangements can be formed with the letters in

$MOUSE$ if the consonants must be on the two ends?

1. $6$
2. $60$
3. $120$
4. $12$

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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 $(n=r)$

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

Find the permutations for the vowels in the middle, and for the consonants on both ends, then multiply them.

First, count the arrangements for the

$3$ vowels

$OUE$ using permutation

$n=r=3$

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

$_\color{purple}{3}P_{\color{purple}{3}}$	$=$	$3!$	Substitute the value of $n$

	$=$	$3\cdot2\cdot1$
	$=$	$6$

There are

$6$ ways of arranging the vowels

$OUE$

Next, count the possible arrangements for the

$2$ consonants

$M$ and

$S$

$n=r=2$

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

$M$ and

$S$

Finally, multiply the two solved permutations

vowels

$=6$

consonants

$=2$

$6\times2$

$=$

$12$

Therefore, there are $12$ ways of arranging the letters in

$MOUSE$ if the two consonants must be on both ends

$12$

Question 7 of 7

How many arrangements can be made with

$MOUSE$ if

$MS$ must always be together?

(48)

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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 $(n=r)$

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

Multiply the number of arrangements if

$MS$ is treated as a single letter to the number or arrangements for letters

$M$ and

$S$ .

First, treat

$MS$ as a single letter. This leaves us with only $4$ letters $(r)$ to be arranged in $4$ positions $(n)$

$n=r=4$

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

$OUE$ and

$MS$

Next, count all possible arrangements for

$MS$ . 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

$M$ and

$S$

Finally, multiply the two solved permutations

_$4$

$P$ _$4$

$=24$

_$2$

$P$ _$2$

$=2$

$24\cdot2$

$=$

$48$

Therefore, there are $48$ ways of arranging

$MOUSE$ if

$MS$ must always be together

$48$

Permutations with Restrictions 2

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Quiz summary

Information

1. Question

Hint

Great Work!

Fundamental Counting Principle

2. Question

Hint

Keep Going!

Fundamental Counting Principle

3. Question

Hint

Excellent!

Permutation Formula if $(n=r)$

4. Question

Hint

Well Done!

Permutation Formula if $(n=r)$

5. Question

Hint

Correct!

Permutation Formula if $(n=r)$

6. Question

Hint

Fantastic!

Permutation Formula if $(n=r)$

7. Question

Hint

Nice Job!

Permutation Formula if $(n=r)$

Quizzes

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$ $n$	Fundamental Counting Principle
	$=$ $=$	$3$ $3$ $\times$ $times$ $6$ $6$ $\times$ $times$ $5$ $5$ $\times$ $times$ $4$ $4$
	$=$ $=$	$360$ $360$

number of ways	$=$ $=$	$m$ $m$ $\times$ $times$ $n$	Fundamental Counting Principle
	$=$	$3$ $times$ $7^3$
	$=$	$1029$

Permutations with Restrictions 2

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Quiz summary

Information

1. Question

Hint

Great Work!

Fundamental Counting Principle

2. Question

Hint

Keep Going!

Fundamental Counting Principle

3. Question

Hint

Excellent!

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

4. Question

Hint

Well Done!

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

5. Question

Hint

Correct!

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

6. Question

Hint

Fantastic!

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

7. Question

Hint

Nice Job!

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 $(n=r)$

Permutation Formula if $(n=r)$