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Permutations with Restrictions 2Permutations with Restrictions 2
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Question 1 of 7
1. Question
How many fourdigit numbers can be formed WITHOUT repetition using the numbers: `1``7`?Hint
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Fundamental Counting Principle
number of ways `=``m``times``n`Count the numbers that can be chosen for each place valueFirst digit:From `1``7`, we can choose `5,6` and `7` to make sure the number is greater than `5000`. Hence, we have `3` choices`=``3`Second digit:From `1``7`, one has already been chosen for the First digit. Hence, we are left with `6` choices`=``6`Third digit:From `1``7`, two has already been chosen for the First and Second digits. Hence, we are left with `5` choices`=``5`Fourth digit:From `1``7`, three has already been chosen for the First, Second and Third digits. Hence, we are left with `4` choices`=``4`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `3``times``6``times``5``times``4` `=` `360` There are `360` ways of forming fourdigit numbers greater than `5000` using `1``7` without repetition`360` 
Question 2 of 7
2. Question
How many fourdigit numbers can be formed WITH repetition using the numbers: `1``7`?Hint
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Fundamental Counting Principle
number of ways `=``m``times``n`Count the numbers that can be chosen for each place valueRemember that the numbers can be repeatedFirst digit:From `1``7`, we can choose `5,6` and `7` to make sure the number is greater than `5000`. Hence, we have `3` choices`=``3`Second to Fourth digit:From `1``7`, we can choose any number and have it repeated. Hence, the other three digits have `7` choices each`=``7`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `3``times``7^3` `=` `1029` There are `1029` ways of forming fourdigit numbers greater than `5000` using `1``7` with repetition`1029` 
Question 3 of 7
3. Question
How many arrangements can be made with `ABCDEFG` if `A` must be placed in the middle?Hint
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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
4. Question
How many arrangements can be made with `ABCDEFG` if `FG` must always be together?Hint
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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
5. Question
How many arrangements can be made with `ABCDEFG` if `F` and `G` must not be together?Hint
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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 togetherFirst, 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 togetherFinally, subtract the arrangement where `FG` must be together from the total number of arrangementsTotal arrangements`=5040`Arrangements where `FG` are together`=1440`$$50401440$$ `=` $$3600$$ Therefore, there are `3600` ways of arranging `ABCDEFG` if `F` and `G` must not be together`3600` 
Question 6 of 7
6. Question
How many arrangements can be formed with the letters in `MOUSE` if the consonants must be on the two ends?Hint
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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 permutationsvowels`=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
7. Question
How many arrangements can be made with `MOUSE` if `MS` must always be together? (48)
Hint
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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`
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