6 periods and 5 subjects
Question : There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ? I am confused.
Initially this appeared as a simple question. By goggling a bit, I am stuck with two answers. Different sites gives different answers and am unable to decide which is right.
Approach 1 (Source)
we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
Approach 2 (Source)
subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
Is any of these approach is right or is the answer different?
combinatorics
asked Jun 1 '15 at 14:03
Kiran
3,2141633
add a comment |
Question : There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ? I am confused.
Initially this appeared as a simple question. By goggling a bit, I am stuck with two answers. Different sites gives different answers and am unable to decide which is right.
Approach 1 (Source)
we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
Approach 2 (Source)
subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
Is any of these approach is right or is the answer different?
combinatorics
asked Jun 1 '15 at 14:03
Kiran
3,2141633
add a comment |
Question : There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ? I am confused.
Initially this appeared as a simple question. By goggling a bit, I am stuck with two answers. Different sites gives different answers and am unable to decide which is right.
Approach 1 (Source)
we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
Approach 2 (Source)
subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
Is any of these approach is right or is the answer different?
combinatorics
asked Jun 1 '15 at 14:03
Kiran
3,2141633
Question : There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ? I am confused.
Initially this appeared as a simple question. By goggling a bit, I am stuck with two answers. Different sites gives different answers and am unable to decide which is right.
Approach 1 (Source)
we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
Approach 2 (Source)
subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
Is any of these approach is right or is the answer different?
combinatorics
combinatorics
asked Jun 1 '15 at 14:03
Kiran
3,2141633
asked Jun 1 '15 at 14:03
Kiran
3,2141633
edited Jun 1 '15 at 14:12
asked Jun 1 '15 at 14:03
Kiran
3,2141633
asked Jun 1 '15 at 14:03
Kiran
3,2141633
asked Jun 1 '15 at 14:03
Kiran
3,2141633
3,2141633
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Approach 1 is incorrect. It arranges any five of the courses into one period each, then assigns the left over course to some period. It double counts each arrangement by having each of the two courses taught at the same time included in the original five. So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period. Approach 2 is the same, but acknowledges the double counting in the division by $2!$
I cannot follow the argument in the paragraph starting "Alternatively"
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
add a comment |
As Ross Millikan explains, the first approach overcounts and the second approach is a correct version.
An alternative way to do the problem is to notice that there must be exactly one pair of periods with the same subject. You can choose this pair in ${}^6mathrm C_2=15$ ways. Once you have made this choice, you have five subjects and five places to put them (one of the "places" being the two periods you've paired off). So there are $5!$ ways to do this. Therefore the answer is ${}^6mathrm C_2times 5!=1800$
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
add a comment |
5 subjects on 5 periods
selecting a place for remaining one subject
Final answer
Over-counting check
answered Dec 20 '17 at 6:48
Mani kandan
11
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
add a comment |
i am a student of class 10 and even i had the same question in my book.
But neither of your answers are correct. The exact answer in 600. I can explain.
_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go!
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
2
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
add a comment |
protected by Community♦ Dec 7 at 17:12
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Approach 1 is incorrect. It arranges any five of the courses into one period each, then assigns the left over course to some period. It double counts each arrangement by having each of the two courses taught at the same time included in the original five. So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period. Approach 2 is the same, but acknowledges the double counting in the division by $2!$
I cannot follow the argument in the paragraph starting "Alternatively"
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
add a comment |
Approach 1 is incorrect. It arranges any five of the courses into one period each, then assigns the left over course to some period. It double counts each arrangement by having each of the two courses taught at the same time included in the original five. So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period. Approach 2 is the same, but acknowledges the double counting in the division by $2!$
I cannot follow the argument in the paragraph starting "Alternatively"
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
add a comment |
Approach 1 is incorrect. It arranges any five of the courses into one period each, then assigns the left over course to some period. It double counts each arrangement by having each of the two courses taught at the same time included in the original five. So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period. Approach 2 is the same, but acknowledges the double counting in the division by $2!$
I cannot follow the argument in the paragraph starting "Alternatively"
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
Approach 1 is incorrect. It arranges any five of the courses into one period each, then assigns the left over course to some period. It double counts each arrangement by having each of the two courses taught at the same time included in the original five. So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period. Approach 2 is the same, but acknowledges the double counting in the division by $2!$
I cannot follow the argument in the paragraph starting "Alternatively"
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
answered Jun 1 '15 at 14:28
Ross Millikan
291k23196370
291k23196370
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
add a comment |
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
ok. so 3600 is incorrect and 1800 is the answer?
– Kiran
Jun 1 '15 at 14:50
add a comment |
As Ross Millikan explains, the first approach overcounts and the second approach is a correct version.
An alternative way to do the problem is to notice that there must be exactly one pair of periods with the same subject. You can choose this pair in ${}^6mathrm C_2=15$ ways. Once you have made this choice, you have five subjects and five places to put them (one of the "places" being the two periods you've paired off). So there are $5!$ ways to do this. Therefore the answer is ${}^6mathrm C_2times 5!=1800$
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
add a comment |
As Ross Millikan explains, the first approach overcounts and the second approach is a correct version.
An alternative way to do the problem is to notice that there must be exactly one pair of periods with the same subject. You can choose this pair in ${}^6mathrm C_2=15$ ways. Once you have made this choice, you have five subjects and five places to put them (one of the "places" being the two periods you've paired off). So there are $5!$ ways to do this. Therefore the answer is ${}^6mathrm C_2times 5!=1800$
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
add a comment |
As Ross Millikan explains, the first approach overcounts and the second approach is a correct version.
An alternative way to do the problem is to notice that there must be exactly one pair of periods with the same subject. You can choose this pair in ${}^6mathrm C_2=15$ ways. Once you have made this choice, you have five subjects and five places to put them (one of the "places" being the two periods you've paired off). So there are $5!$ ways to do this. Therefore the answer is ${}^6mathrm C_2times 5!=1800$
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
As Ross Millikan explains, the first approach overcounts and the second approach is a correct version.
An alternative way to do the problem is to notice that there must be exactly one pair of periods with the same subject. You can choose this pair in ${}^6mathrm C_2=15$ ways. Once you have made this choice, you have five subjects and five places to put them (one of the "places" being the two periods you've paired off). So there are $5!$ ways to do this. Therefore the answer is ${}^6mathrm C_2times 5!=1800$
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
answered Dec 4 '17 at 11:40
Especially Lime
21.6k22758
21.6k22758
add a comment |
add a comment |
5 subjects on 5 periods
selecting a place for remaining one subject
Final answer
Over-counting check
answered Dec 20 '17 at 6:48
Mani kandan
11
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
add a comment |
5 subjects on 5 periods
selecting a place for remaining one subject
Final answer
Over-counting check
answered Dec 20 '17 at 6:48
Mani kandan
11
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
add a comment |
5 subjects on 5 periods
selecting a place for remaining one subject
Final answer
Over-counting check
answered Dec 20 '17 at 6:48
Mani kandan
11
5 subjects on 5 periods
selecting a place for remaining one subject
Final answer
Over-counting check
answered Dec 20 '17 at 6:48
Mani kandan
11
answered Dec 20 '17 at 6:48
Mani kandan
11
answered Dec 20 '17 at 6:48
Mani kandan
11
answered Dec 20 '17 at 6:48
Mani kandan
11
11
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
add a comment |
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
It is good that you have produced valuable information in answering this question. I think your efforts would be more well-received if you were to use MathJAX and put the content of your answer directly in this space, rather than linking to some outside location. You might consider reading math.stackexchange.com/help/notation as a start.
– abiessu
Dec 20 '17 at 7:05
add a comment |
i am a student of class 10 and even i had the same question in my book.
But neither of your answers are correct. The exact answer in 600. I can explain.
_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go!
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
2
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
add a comment |
i am a student of class 10 and even i had the same question in my book.
But neither of your answers are correct. The exact answer in 600. I can explain.
_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go!
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
2
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
add a comment |
i am a student of class 10 and even i had the same question in my book.
But neither of your answers are correct. The exact answer in 600. I can explain.
_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go!
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
i am a student of class 10 and even i had the same question in my book.
But neither of your answers are correct. The exact answer in 600. I can explain.
_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go!
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
answered Dec 4 '17 at 11:19
Shreeya Kamathh
1
1
2
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
add a comment |
2
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
2
2
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
This is incorrect, since it assumes that the first five periods all have different subjects, which need not be true.
– Especially Lime
Dec 4 '17 at 11:34
add a comment |
protected by Community♦ Dec 7 at 17:12
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
