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Permutations with Restrictions 1Permutations with Restrictions 1
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Question 1 of 7
1. Question
How many numbers greater than `400` can be formed using the numbers: `1,4,7` and `9`?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 total arrangements for both `4`digit and `3`digit numbers then add them.First, count the `4`digit numbers greater than `400` using permutationSince `400` is only a `3`digit number, all `4`digit arrangements using `1,4,7` and `9` are greater than `400``n=r=4`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$n!$$ Permutation Formula if `(n=r)` $$_\color{purple}{4}P_{\color{purple}{4}}$$ `=` $$ 4! $$ Substitute the value of `n` `=` $$4\cdot3\cdot2\cdot1$$ `=` $$24$$ There are `24` ways of forming `4`digit numbers greater than `400`, using `1,4,7` and `9`Next, count the `3`digit numbers greater than `400` by using the Fundamental Counting PrincipleStart by listing down and counting the options for each place valueFirst digit:From `1,4,7,9`, we can only choose `4,7` and `9` because if the `3`digit number starts with `1`, it would be less than `400``=``3`Second digit:From `1,4,7,9`, one has already been chosen for the First digit. Hence, we are left with `3` choices`=``3`Third digit:From `1,4,7,9`, two has already been chosen for the First and Second digits. Hence, we are left with `2` choices`=``2`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``3``times``2` `=` `18` There are `18` ways of forming `3`digit numbers greater than `400`, using `1,4,7` and `9`Finally, add the two solved permutations`4`digit numbers`=24``3`digit numbers`=18`$$24+18$$ `=` $$42$$ Therefore, there are `42` ways of forming a number greater than `400` using the numbers `1,4,7` and `9`.`42` 
Question 2 of 7
2. Question
How many numbers greater than `600` can be formed using the numbers: `2,6,4` and `9`?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 total arrangements for both `4`digit and `3`digit numbers then add them.First, count the `4`digit numbers greater than `600` using permutationSince `600` is only a `3`digit number, all `4`digit arrangements using `2,6,4` and `9` are greater than `600``n=r=4`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$n!$$ Permutation Formula if `(n=r)` $$_\color{purple}{4}P_{\color{purple}{4}}$$ `=` $$ 4! $$ Substitute the value of `n` `=` $$4\cdot3\cdot2\cdot1$$ `=` $$24$$ There are `24` ways of forming `4`digit numbers greater than `600`, using `2,6,4` and `9`Next, count the `3`digit numbers greater than `600` by using the Fundamental Counting PrincipleStart by listing down and counting the options for each place valueFirst digit:From `2,6,4,9`, we can only choose `6` and `9` because if the `3`digit number starts with `2` or `4`, it would be less than `600``=``2`Second digit:From `2,6,4,9`, one has already been chosen for the First digit. Hence, we are left with `3` choices`=``3`Third digit:From `2,6,4,9`, two has already been chosen for the First and Second digits. Hence, we are left with `2` choices`=``2`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `2``times``3``times``2` `=` `12` There are `12` ways of forming `3`digit numbers greater than `600`, using `2,6,4` and `9`Finally, add the two solved permutations`4`digit numbers`=24``3`digit numbers`=12`$$24+12$$ `=` $$36$$ Therefore, there are `36` ways of forming a number greater than `600` using the numbers `2,6,4` and `9`.`36` 
Question 3 of 7
3. Question
How many numbers between `7000` to `8000` can be formed using the numbers: `6,7,8` and `9`? (6)
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Fundamental Counting Principle
number of ways `=``m``times``n`Remember that the number has to be between `7000` to `8000`, which means the number has to have four digitsFirst digit:From `6,7,8,9`, we can only choose `7` because if the `4`digit number starts with `6`, it would be less than `7000` and if it starts with `8` or `9`, it would be greater than `8000``=``1`Second digit:From `6,7,8,9`, the number `7` has already been chosen for the First digit. Hence, we are left with `3` choices`=``3`Third digit:From `6,7,8,9`, two has already been chosen for the First and Second digits. Hence, we are left with `2` choices`=``2`Fourth digit:From `6,7,8,9`, three has already been chosen for the First, Second and Third digits. Hence, we are left with `1` choice`=``1`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `1``times``3``times``2``times``1` `=` `6` There are `6` ways of forming numbers between `7000` to `8000`, using `6,7,8` and `9``6` 
Question 4 of 7
4. Question
How many numbers between `7000` to `10 000` can be formed using the numbers: `6,7,8` and `9`? (18)
Hint
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Fundamental Counting Principle
number of ways `=``m``times``n`Remember that the number has to be between `7000` to `10 000`, which means the number has to have four digitsFirst digit:From `6,7,8,9`, we can choose `7,8` and `9` because if the `4`digit number starts with `6`, it would be less than `7000``=``3`Second digit:From `6,7,8,9`, one has already been chosen for the First digit. Hence, we are left with `3` choices`=``3`Third digit:From `6,7,8,9`, two has already been chosen for the First and Second digits. Hence, we are left with `2` choices`=``2`Fourth digit:From `6,7,8,9`, three has already been chosen for the First, Second and Third digits. Hence, we are left with `1` choice`=``1`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``3``times``2``times``1` `=` `18` There are `18` ways of forming numbers between `7000` to `10 000`, using `6,7,8` and `9``18` 
Question 5 of 7
5. Question
How many fivedigit even numbers can be formed using the numbers: `2,3,4,5` and `6`? (72)
Hint
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Fundamental Counting Principle
number of ways `=``m``times``n`Count the numbers that can be chosen for each place valueSince we are looking for even numbers, start with the last or fifth digit to make sure it is evenFifth digit:From `2,3,4,5,6`, we can choose `2,4` and `6` since they are even. Hence, we have `3` choices`=``3`First digit:From `2,3,4,5,6`, one has already been chosen for the Fifth digit. Hence, we are left with `4` choices`=``4`Second digit:From `2,3,4,5,6`, two has already been chosen for the Fifth and First digit. Hence, we are left with `3` choices`=``3`Third digit:From `2,3,4,5,6`, three has already been chosen for the Fifth, First and Second digits. Hence, we are left with `2` choices`=``2`Fourth digit:From `2,3,4,5,6`, four has already been chosen for the Fifth, First, Second and Third digits. Hence, we are left with `1` choice`=``1`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``4``times``3``times``2``times``1` `=` `72` There are `72` ways of forming fivedigit even numbers using `2,3,4,5` and `6``72` 
Question 6 of 7
6. Question
How many sixdigit numbers can be formed WITHOUT repetition using the numbers: `0``9`?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 `0``9`, we can choose all digits except `0` to make sure that we keep the number at six digits. Hence, we have `9` choices`=``9`Second digit:From `0``9`, one has already been chosen for the First digit. Hence, we are left with `9` choices`=``9`Third digit:From `0``9`, two has already been chosen for the First and Second digits. Hence, we are left with `8` choices`=``8`Fourth digit:From `0``9`, three has already been chosen for the First, Second and Third digits. Hence, we are left with `7` choices`=``7`Fifth digit:From `0``9`, four has already been chosen for the First, Second, Third and Fourth digits. Hence, are left with `6` choices`=``6`Sixth digit:From `0``9`, five has already been chosen for the First, Second, Third, Fourth and Fifth digits. Hence, are left with `5` choices`=``5`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `9``times``9``times``8``times``7``times``6``times``5` `=` `136 080` There are `136 080` ways of forming sixdigit numbers using `0``9` without repetition`136 080` 
Question 7 of 7
7. Question
How many sixdigit numbers can be formed WITH repetition using the numbers: `0``9`?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 `0``9`, we can choose all digits except `0` to make sure that we keep the number at six digits. Hence, we have `9` choices`=``9`Second to Sixth digits:From `0``9`, we can choose any digit and have it repeated. Hence, the other five digits have `10` choices each`=``10`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `9``times``10^5` `=` `9times100 000` `=` `900 000` There are `900 000` ways of forming sixdigit numbers using `0``9` with repetition`900 000`
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