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Largest number of students who can attend special lesson [closed]
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A teacher provides four special lessons, one each in Maths, Music,
English and Science, for some of the children in her class. For the
students in these special lessons:
• there are exactly 3 children in each lesson.
• each pair of students attends at least one special lesson together.
What is the largest number of students who can attend these special
lessons?
Can someone help me solve this problem?
combinatorics combinations
asked Jul 22 at 20:56
geetha
585
closed as off-topic by amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel Jul 23 at 2:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel
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up vote
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A teacher provides four special lessons, one each in Maths, Music,
English and Science, for some of the children in her class. For the
students in these special lessons:
• there are exactly 3 children in each lesson.
• each pair of students attends at least one special lesson together.
What is the largest number of students who can attend these special
lessons?
Can someone help me solve this problem?
combinatorics combinations
asked Jul 22 at 20:56
geetha
585
closed as off-topic by amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel Jul 23 at 2:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel
2
Welcome to MSE. It is expected that you show your working to show how far you have got in solving the problem by yourself. Please, if you can, edit your question with any ideas you have, no matter how small. Then people can see how stuck you are, and on which parts. Do this in the body of the question and don't leave it for the comment section.
– Daniel Buck
Jul 22 at 21:31
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up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
A teacher provides four special lessons, one each in Maths, Music,
English and Science, for some of the children in her class. For the
students in these special lessons:
• there are exactly 3 children in each lesson.
• each pair of students attends at least one special lesson together.
What is the largest number of students who can attend these special
lessons?
Can someone help me solve this problem?
combinatorics combinations
asked Jul 22 at 20:56
geetha
585
A teacher provides four special lessons, one each in Maths, Music,
English and Science, for some of the children in her class. For the
students in these special lessons:
• there are exactly 3 children in each lesson.
• each pair of students attends at least one special lesson together.
What is the largest number of students who can attend these special
lessons?
Can someone help me solve this problem?
combinatorics combinations
asked Jul 22 at 20:56
geetha
585
asked Jul 22 at 20:56
geetha
585
asked Jul 22 at 20:56
geetha
585
asked Jul 22 at 20:56
geetha
585
585
closed as off-topic by amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel Jul 23 at 2:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel
closed as off-topic by amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel Jul 23 at 2:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, user21820, Parcly Taxel
2
Welcome to MSE. It is expected that you show your working to show how far you have got in solving the problem by yourself. Please, if you can, edit your question with any ideas you have, no matter how small. Then people can see how stuck you are, and on which parts. Do this in the body of the question and don't leave it for the comment section.
– Daniel Buck
Jul 22 at 21:31
add a comment |Â
2
Welcome to MSE. It is expected that you show your working to show how far you have got in solving the problem by yourself. Please, if you can, edit your question with any ideas you have, no matter how small. Then people can see how stuck you are, and on which parts. Do this in the body of the question and don't leave it for the comment section.
– Daniel Buck
Jul 22 at 21:31
2
2
Welcome to MSE. It is expected that you show your working to show how far you have got in solving the problem by yourself. Please, if you can, edit your question with any ideas you have, no matter how small. Then people can see how stuck you are, and on which parts. Do this in the body of the question and don't leave it for the comment section.
– Daniel Buck
Jul 22 at 21:31
Welcome to MSE. It is expected that you show your working to show how far you have got in solving the problem by yourself. Please, if you can, edit your question with any ideas you have, no matter how small. Then people can see how stuck you are, and on which parts. Do this in the body of the question and don't leave it for the comment section.
– Daniel Buck
Jul 22 at 21:31
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2 Answers
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Consider the number of possible pairs in the lessons, each class has $3$ pairs and the there are $4$ classes, so that makes $12$ possible pairs.
If there are $5$ students there are $10$ pairs, if there are $6$ students there are $15$ pairs (too many!), so the largest number of students is possibly $5$, now can all $10$ pairs be realised in the four classes ...
The answer is yes ...
$a,b,c ,a,d,e,b,c,d,b,c,e $
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
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With six kids, there are $binom62=15$ different pairs. Each lesson hosts $binom32=3$ pairs, and there are only four lessons, so at least three pairs will never meet. Therefore, six is too many.
Can you make it happen with five? Try to build a complete graph on five vertices, i.e., a pentagon with a star inscribed, out of four triangles, overlapping allowed (indeed, necessary).
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Consider the number of possible pairs in the lessons, each class has $3$ pairs and the there are $4$ classes, so that makes $12$ possible pairs.
If there are $5$ students there are $10$ pairs, if there are $6$ students there are $15$ pairs (too many!), so the largest number of students is possibly $5$, now can all $10$ pairs be realised in the four classes ...
The answer is yes ...
$a,b,c ,a,d,e,b,c,d,b,c,e $
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
add a comment |Â
up vote
1
down vote
Consider the number of possible pairs in the lessons, each class has $3$ pairs and the there are $4$ classes, so that makes $12$ possible pairs.
If there are $5$ students there are $10$ pairs, if there are $6$ students there are $15$ pairs (too many!), so the largest number of students is possibly $5$, now can all $10$ pairs be realised in the four classes ...
The answer is yes ...
$a,b,c ,a,d,e,b,c,d,b,c,e $
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider the number of possible pairs in the lessons, each class has $3$ pairs and the there are $4$ classes, so that makes $12$ possible pairs.
If there are $5$ students there are $10$ pairs, if there are $6$ students there are $15$ pairs (too many!), so the largest number of students is possibly $5$, now can all $10$ pairs be realised in the four classes ...
The answer is yes ...
$a,b,c ,a,d,e,b,c,d,b,c,e $
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
Consider the number of possible pairs in the lessons, each class has $3$ pairs and the there are $4$ classes, so that makes $12$ possible pairs.
If there are $5$ students there are $10$ pairs, if there are $6$ students there are $15$ pairs (too many!), so the largest number of students is possibly $5$, now can all $10$ pairs be realised in the four classes ...
The answer is yes ...
$a,b,c ,a,d,e,b,c,d,b,c,e $
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
answered Jul 22 at 21:10
Donald Splutterwit
21.3k21243
21.3k21243
add a comment |Â
add a comment |Â
up vote
0
down vote
With six kids, there are $binom62=15$ different pairs. Each lesson hosts $binom32=3$ pairs, and there are only four lessons, so at least three pairs will never meet. Therefore, six is too many.
Can you make it happen with five? Try to build a complete graph on five vertices, i.e., a pentagon with a star inscribed, out of four triangles, overlapping allowed (indeed, necessary).
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
add a comment |Â
up vote
0
down vote
With six kids, there are $binom62=15$ different pairs. Each lesson hosts $binom32=3$ pairs, and there are only four lessons, so at least three pairs will never meet. Therefore, six is too many.
Can you make it happen with five? Try to build a complete graph on five vertices, i.e., a pentagon with a star inscribed, out of four triangles, overlapping allowed (indeed, necessary).
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
add a comment |Â
up vote
0
down vote
up vote
0
down vote
With six kids, there are $binom62=15$ different pairs. Each lesson hosts $binom32=3$ pairs, and there are only four lessons, so at least three pairs will never meet. Therefore, six is too many.
Can you make it happen with five? Try to build a complete graph on five vertices, i.e., a pentagon with a star inscribed, out of four triangles, overlapping allowed (indeed, necessary).
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
With six kids, there are $binom62=15$ different pairs. Each lesson hosts $binom32=3$ pairs, and there are only four lessons, so at least three pairs will never meet. Therefore, six is too many.
Can you make it happen with five? Try to build a complete graph on five vertices, i.e., a pentagon with a star inscribed, out of four triangles, overlapping allowed (indeed, necessary).
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
answered Jul 22 at 21:20
G Tony Jacobs
25.6k43483
25.6k43483
add a comment |Â
add a comment |Â
2
Welcome to MSE. It is expected that you show your working to show how far you have got in solving the problem by yourself. Please, if you can, edit your question with any ideas you have, no matter how small. Then people can see how stuck you are, and on which parts. Do this in the body of the question and don't leave it for the comment section.
– Daniel Buck
Jul 22 at 21:31