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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?- 1.
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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)
nPn=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 400n=r=4nPn = n! Permutation Formula if (n=r) 4P4 = 4! Substitute the value of n = 4⋅3⋅2⋅1 = 24 There are 24 ways of forming 4-digit numbers greater than 400, using 1,4,7 and 9Next, 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=3Second digit:From 1,4,7,9, one has already been chosen for the First digit. Hence, we are left with 3 choices=3Third digit:From 1,4,7,9, two has already been chosen for the First and Second digits. Hence, we are left with 2 choices=2Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 3×3×2 = 18 There are 18 ways of forming 3-digit numbers greater than 400, using 1,4,7 and 9Finally, add the two solved permutations4-digit numbers=243-digit numbers=1824+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?-
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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)
nPn=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 600n=r=4nPn = n! Permutation Formula if (n=r) 4P4 = 4! Substitute the value of n = 4⋅3⋅2⋅1 = 24 There are 24 ways of forming 4-digit numbers greater than 600, using 2,6,4 and 9Next, 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=2Second digit:From 2,6,4,9, one has already been chosen for the First digit. Hence, we are left with 3 choices=3Third digit:From 2,6,4,9, two has already been chosen for the First and Second digits. Hence, we are left with 2 choices=2Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 2×3×2 = 12 There are 12 ways of forming 3-digit numbers greater than 600, using 2,6,4 and 9Finally, add the two solved permutations4-digit numbers=243-digit numbers=1224+12 = 36 Therefore, there are 36 ways of forming a number greater than 600 using the numbers 2,6,4 and 9.36 -
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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×nRemember 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=1Second digit:From 6,7,8,9, the number 7 has already been chosen for the First digit. Hence, we are left with 3 choices=3Third digit:From 6,7,8,9, two has already been chosen for the First and Second digits. Hence, we are left with 2 choices=2Fourth 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=1Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 1×3×2×1 = 6 There are 6 ways of forming numbers between 7000 to 8000, using 6,7,8 and 96 -
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)
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Fundamental Counting Principle
number of ways =m×nRemember 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=3Second digit:From 6,7,8,9, one has already been chosen for the First digit. Hence, we are left with 3 choices=3Third digit:From 6,7,8,9, two has already been chosen for the First and Second digits. Hence, we are left with 2 choices=2Fourth 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=1Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 3×3×2×1 = 18 There are 18 ways of forming numbers between 7000 to 10 000, using 6,7,8 and 918 -
Question 5 of 7
5. Question
How many five-digit even numbers can be formed using the numbers: 2,3,4,5 and 6?- (72)
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Fundamental Counting Principle
number of ways =m×nCount 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=3First digit:From 2,3,4,5,6, one has already been chosen for the Fifth digit. Hence, we are left with 4 choices=4Second 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=3Third 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=2Fourth 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=1Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 3×4×3×2×1 = 72 There are 72 ways of forming five-digit even numbers using 2,3,4,5 and 672 -
Question 6 of 7
6. Question
How many six-digit numbers can be formed WITHOUT repetition using the numbers: 0-9?-
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Fundamental Counting Principle
number of ways =m×nCount 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=9Second digit:From 0-9, one has already been chosen for the First digit. Hence, we are left with 9 choices=9Third digit:From 0-9, two has already been chosen for the First and Second digits. Hence, we are left with 8 choices=8Fourth digit:From 0-9, three has already been chosen for the First, Second and Third digits. Hence, we are left with 7 choices=7Fifth digit:From 0-9, four has already been chosen for the First, Second, Third and Fourth digits. Hence, are left with 6 choices=6Sixth digit:From 0-9, five has already been chosen for the First, Second, Third, Fourth and Fifth digits. Hence, are left with 5 choices=5Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 9×9×8×7×6×5 = 136 080 There are 136 080 ways of forming six-digit numbers using 0-9 without repetition136 080 -
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Question 7 of 7
7. Question
How many six-digit numbers can be formed WITH repetition using the numbers: 0-9?-
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Fundamental Counting Principle
number of ways =m×nCount 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=9Second to Sixth digits:From 0-9, we can choose any digit and have it repeated. Hence, the other five digits have 10 choices each=10Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways = m×n Fundamental Counting Principle = 9×105 = 9×100 000 = 900 000 There are 900 000 ways of forming six-digit numbers using 0-9 with repetition900 000 -
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Quizzes
- Factorial Notation
- Fundamental Counting Principle 1
- Fundamental Counting Principle 2
- Fundamental Counting Principle 3
- Combinations 1
- Combinations 2
- Combinations with Restrictions 1
- Combinations with Restrictions 2
- Combinations with Probability
- Basic Permutations 1
- Basic Permutations 2
- Basic Permutations 3
- Permutation Problems 1
- Permutation Problems 2
- Permutations with Repetitions 1
- Permutations with Repetitions 2
- Permutations with Restrictions 1
- Permutations with Restrictions 2
- Permutations with Restrictions 3
- Permutations with Restrictions 4