How to approach? - to find the number of visits that the teacher makes and a student makes…
$begingroup$
The teacher has 25 students in her class. She takes 5 of them at a time, to a park as often she can, without taking the same 5 kids more than once. Find the number of visits, the teacher makes to the park and also the number of visits every kid makes.
Attempt
So, I tried to do this and came to a thought that this question means that the same lot can't go to park twice or more. But, any number of students less than 5 can go more than 1 time being a different lot. But, I facing problem on accounting this. It is coming to be very big number which is I think is absurd.
Also the answer key has the answers $25_{C_5}$ & $24_{C_4}$ respectively which is very less as compared to my answer which is more than 7 digit figure.
combinatorics combinations
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
$endgroup$
add a comment |
$begingroup$
The teacher has 25 students in her class. She takes 5 of them at a time, to a park as often she can, without taking the same 5 kids more than once. Find the number of visits, the teacher makes to the park and also the number of visits every kid makes.
Attempt
So, I tried to do this and came to a thought that this question means that the same lot can't go to park twice or more. But, any number of students less than 5 can go more than 1 time being a different lot. But, I facing problem on accounting this. It is coming to be very big number which is I think is absurd.
Also the answer key has the answers $25_{C_5}$ & $24_{C_4}$ respectively which is very less as compared to my answer which is more than 7 digit figure.
combinatorics combinations
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
$endgroup$
1
$begingroup$
The number of visit is simply $binom{25}{5}$. Since by symmetry every student goes the same number, each student goes 1/5 of total number of visits.
$endgroup$
– ablmf
Dec 24 '18 at 0:46
$begingroup$
@ablmf But, that's the problem I am not able to visualize the symmetry.
$endgroup$
– jayant98
Dec 24 '18 at 0:49
$begingroup$
A given student must visit the park with four of the other twenty-four students.
$endgroup$
– N. F. Taussig
Dec 24 '18 at 1:03
$begingroup$
@Taussig I am sorry. But, I am not able to understand it still. Maybe it is due to language problem (I have started learning English from about 10 months). If you may, please present an example. It may clear my thought. Thanks for your effort.
$endgroup$
– jayant98
Dec 24 '18 at 1:09
1
$begingroup$
By definition $^{25}C_5$ means the number of ways of choosing 5 from 25 which is what the question is asking. The symmetry in this case works because there is nothing to distinguish one student from another so the requirements and number of ways each student can be matched with 4 other students is identical for all 25. Symmetry would be broken if say, of the 5 students at least one has to be a girl. This would eliminate all boy groups of 5 and hence reduce the number of visits by the boys compared to the girls.
$endgroup$
– Phil H
Dec 24 '18 at 1:39
add a comment |
$begingroup$
The teacher has 25 students in her class. She takes 5 of them at a time, to a park as often she can, without taking the same 5 kids more than once. Find the number of visits, the teacher makes to the park and also the number of visits every kid makes.
Attempt
So, I tried to do this and came to a thought that this question means that the same lot can't go to park twice or more. But, any number of students less than 5 can go more than 1 time being a different lot. But, I facing problem on accounting this. It is coming to be very big number which is I think is absurd.
Also the answer key has the answers $25_{C_5}$ & $24_{C_4}$ respectively which is very less as compared to my answer which is more than 7 digit figure.
combinatorics combinations
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
$endgroup$
The teacher has 25 students in her class. She takes 5 of them at a time, to a park as often she can, without taking the same 5 kids more than once. Find the number of visits, the teacher makes to the park and also the number of visits every kid makes.
Attempt
So, I tried to do this and came to a thought that this question means that the same lot can't go to park twice or more. But, any number of students less than 5 can go more than 1 time being a different lot. But, I facing problem on accounting this. It is coming to be very big number which is I think is absurd.
Also the answer key has the answers $25_{C_5}$ & $24_{C_4}$ respectively which is very less as compared to my answer which is more than 7 digit figure.
combinatorics combinations
combinatorics combinations
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
edited Dec 24 '18 at 8:41
jayant98
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
asked Dec 24 '18 at 0:37
jayant98jayant98
639217
639217
1
$begingroup$
The number of visit is simply $binom{25}{5}$. Since by symmetry every student goes the same number, each student goes 1/5 of total number of visits.
$endgroup$
– ablmf
Dec 24 '18 at 0:46
$begingroup$
@ablmf But, that's the problem I am not able to visualize the symmetry.
$endgroup$
– jayant98
Dec 24 '18 at 0:49
$begingroup$
A given student must visit the park with four of the other twenty-four students.
$endgroup$
– N. F. Taussig
Dec 24 '18 at 1:03
$begingroup$
@Taussig I am sorry. But, I am not able to understand it still. Maybe it is due to language problem (I have started learning English from about 10 months). If you may, please present an example. It may clear my thought. Thanks for your effort.
$endgroup$
– jayant98
Dec 24 '18 at 1:09
1
$begingroup$
By definition $^{25}C_5$ means the number of ways of choosing 5 from 25 which is what the question is asking. The symmetry in this case works because there is nothing to distinguish one student from another so the requirements and number of ways each student can be matched with 4 other students is identical for all 25. Symmetry would be broken if say, of the 5 students at least one has to be a girl. This would eliminate all boy groups of 5 and hence reduce the number of visits by the boys compared to the girls.
$endgroup$
– Phil H
Dec 24 '18 at 1:39
add a comment |
1
$begingroup$
The number of visit is simply $binom{25}{5}$. Since by symmetry every student goes the same number, each student goes 1/5 of total number of visits.
$endgroup$
– ablmf
Dec 24 '18 at 0:46
$begingroup$
@ablmf But, that's the problem I am not able to visualize the symmetry.
$endgroup$
– jayant98
Dec 24 '18 at 0:49
$begingroup$
A given student must visit the park with four of the other twenty-four students.
$endgroup$
– N. F. Taussig
Dec 24 '18 at 1:03
$begingroup$
@Taussig I am sorry. But, I am not able to understand it still. Maybe it is due to language problem (I have started learning English from about 10 months). If you may, please present an example. It may clear my thought. Thanks for your effort.
$endgroup$
– jayant98
Dec 24 '18 at 1:09
1
$begingroup$
By definition $^{25}C_5$ means the number of ways of choosing 5 from 25 which is what the question is asking. The symmetry in this case works because there is nothing to distinguish one student from another so the requirements and number of ways each student can be matched with 4 other students is identical for all 25. Symmetry would be broken if say, of the 5 students at least one has to be a girl. This would eliminate all boy groups of 5 and hence reduce the number of visits by the boys compared to the girls.
$endgroup$
– Phil H
Dec 24 '18 at 1:39
1
1
$begingroup$
The number of visit is simply $binom{25}{5}$. Since by symmetry every student goes the same number, each student goes 1/5 of total number of visits.
$endgroup$
– ablmf
Dec 24 '18 at 0:46
$begingroup$
The number of visit is simply $binom{25}{5}$. Since by symmetry every student goes the same number, each student goes 1/5 of total number of visits.
$endgroup$
– ablmf
Dec 24 '18 at 0:46
$begingroup$
@ablmf But, that's the problem I am not able to visualize the symmetry.
$endgroup$
– jayant98
Dec 24 '18 at 0:49
$begingroup$
@ablmf But, that's the problem I am not able to visualize the symmetry.
$endgroup$
– jayant98
Dec 24 '18 at 0:49
$begingroup$
A given student must visit the park with four of the other twenty-four students.
$endgroup$
– N. F. Taussig
Dec 24 '18 at 1:03
$begingroup$
A given student must visit the park with four of the other twenty-four students.
$endgroup$
– N. F. Taussig
Dec 24 '18 at 1:03
$begingroup$
@Taussig I am sorry. But, I am not able to understand it still. Maybe it is due to language problem (I have started learning English from about 10 months). If you may, please present an example. It may clear my thought. Thanks for your effort.
$endgroup$
– jayant98
Dec 24 '18 at 1:09
$begingroup$
@Taussig I am sorry. But, I am not able to understand it still. Maybe it is due to language problem (I have started learning English from about 10 months). If you may, please present an example. It may clear my thought. Thanks for your effort.
$endgroup$
– jayant98
Dec 24 '18 at 1:09
1
1
$begingroup$
By definition $^{25}C_5$ means the number of ways of choosing 5 from 25 which is what the question is asking. The symmetry in this case works because there is nothing to distinguish one student from another so the requirements and number of ways each student can be matched with 4 other students is identical for all 25. Symmetry would be broken if say, of the 5 students at least one has to be a girl. This would eliminate all boy groups of 5 and hence reduce the number of visits by the boys compared to the girls.
$endgroup$
– Phil H
Dec 24 '18 at 1:39
$begingroup$
By definition $^{25}C_5$ means the number of ways of choosing 5 from 25 which is what the question is asking. The symmetry in this case works because there is nothing to distinguish one student from another so the requirements and number of ways each student can be matched with 4 other students is identical for all 25. Symmetry would be broken if say, of the 5 students at least one has to be a girl. This would eliminate all boy groups of 5 and hence reduce the number of visits by the boys compared to the girls.
$endgroup$
– Phil H
Dec 24 '18 at 1:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's consider a smaller example. Suppose the teacher has six students, $A, B, C, D, E, F$, and takes three students to the park at the time. There are $binom{6}{3} = 20$ groups she can take. They are
$$
begin{array}{c c}
color{red}{A}, B, C & B, C, D\
color{red}{A}, B, D & B, C, E\
color{red}{A}, B, E & B, C, F\
color{red}{A}, B, F & B, D, E\
color{red}{A}, C, D & B, D, F\
color{red}{A}, C, E & B, E, F\
color{red}{A}, C, F & C, D, E\
color{red}{A}, D, E & C, D, F\
color{red}{A}, D, F & C, E, F\
color{red}{A}, E, F & D, E, F
end{array}
$$
Notice that $A$ is in $binom{5}{2} = 10$ of these groups, once for each of the $binom{5}{2}$ ways the teacher can select two students to go to the park with $A$.
In your problem, there are $binom{25}{5}$ ways for the teacher to select $5$ of the $25$ students to go to the park and $binom{24}{4}$ ways for the teacher to select four other students to go the park with a particular student. Hence, each student is in $binom{24}{4}$ of the $binom{25}{5}$ possible groups the teacher can take to the park.
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
add a comment |
Your Answer
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1 Answer
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$begingroup$
Let's consider a smaller example. Suppose the teacher has six students, $A, B, C, D, E, F$, and takes three students to the park at the time. There are $binom{6}{3} = 20$ groups she can take. They are
$$
begin{array}{c c}
color{red}{A}, B, C & B, C, D\
color{red}{A}, B, D & B, C, E\
color{red}{A}, B, E & B, C, F\
color{red}{A}, B, F & B, D, E\
color{red}{A}, C, D & B, D, F\
color{red}{A}, C, E & B, E, F\
color{red}{A}, C, F & C, D, E\
color{red}{A}, D, E & C, D, F\
color{red}{A}, D, F & C, E, F\
color{red}{A}, E, F & D, E, F
end{array}
$$
Notice that $A$ is in $binom{5}{2} = 10$ of these groups, once for each of the $binom{5}{2}$ ways the teacher can select two students to go to the park with $A$.
In your problem, there are $binom{25}{5}$ ways for the teacher to select $5$ of the $25$ students to go to the park and $binom{24}{4}$ ways for the teacher to select four other students to go the park with a particular student. Hence, each student is in $binom{24}{4}$ of the $binom{25}{5}$ possible groups the teacher can take to the park.
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
add a comment |
$begingroup$
Let's consider a smaller example. Suppose the teacher has six students, $A, B, C, D, E, F$, and takes three students to the park at the time. There are $binom{6}{3} = 20$ groups she can take. They are
$$
begin{array}{c c}
color{red}{A}, B, C & B, C, D\
color{red}{A}, B, D & B, C, E\
color{red}{A}, B, E & B, C, F\
color{red}{A}, B, F & B, D, E\
color{red}{A}, C, D & B, D, F\
color{red}{A}, C, E & B, E, F\
color{red}{A}, C, F & C, D, E\
color{red}{A}, D, E & C, D, F\
color{red}{A}, D, F & C, E, F\
color{red}{A}, E, F & D, E, F
end{array}
$$
Notice that $A$ is in $binom{5}{2} = 10$ of these groups, once for each of the $binom{5}{2}$ ways the teacher can select two students to go to the park with $A$.
In your problem, there are $binom{25}{5}$ ways for the teacher to select $5$ of the $25$ students to go to the park and $binom{24}{4}$ ways for the teacher to select four other students to go the park with a particular student. Hence, each student is in $binom{24}{4}$ of the $binom{25}{5}$ possible groups the teacher can take to the park.
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
add a comment |
$begingroup$
Let's consider a smaller example. Suppose the teacher has six students, $A, B, C, D, E, F$, and takes three students to the park at the time. There are $binom{6}{3} = 20$ groups she can take. They are
$$
begin{array}{c c}
color{red}{A}, B, C & B, C, D\
color{red}{A}, B, D & B, C, E\
color{red}{A}, B, E & B, C, F\
color{red}{A}, B, F & B, D, E\
color{red}{A}, C, D & B, D, F\
color{red}{A}, C, E & B, E, F\
color{red}{A}, C, F & C, D, E\
color{red}{A}, D, E & C, D, F\
color{red}{A}, D, F & C, E, F\
color{red}{A}, E, F & D, E, F
end{array}
$$
Notice that $A$ is in $binom{5}{2} = 10$ of these groups, once for each of the $binom{5}{2}$ ways the teacher can select two students to go to the park with $A$.
In your problem, there are $binom{25}{5}$ ways for the teacher to select $5$ of the $25$ students to go to the park and $binom{24}{4}$ ways for the teacher to select four other students to go the park with a particular student. Hence, each student is in $binom{24}{4}$ of the $binom{25}{5}$ possible groups the teacher can take to the park.
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
Let's consider a smaller example. Suppose the teacher has six students, $A, B, C, D, E, F$, and takes three students to the park at the time. There are $binom{6}{3} = 20$ groups she can take. They are
$$
begin{array}{c c}
color{red}{A}, B, C & B, C, D\
color{red}{A}, B, D & B, C, E\
color{red}{A}, B, E & B, C, F\
color{red}{A}, B, F & B, D, E\
color{red}{A}, C, D & B, D, F\
color{red}{A}, C, E & B, E, F\
color{red}{A}, C, F & C, D, E\
color{red}{A}, D, E & C, D, F\
color{red}{A}, D, F & C, E, F\
color{red}{A}, E, F & D, E, F
end{array}
$$
Notice that $A$ is in $binom{5}{2} = 10$ of these groups, once for each of the $binom{5}{2}$ ways the teacher can select two students to go to the park with $A$.
In your problem, there are $binom{25}{5}$ ways for the teacher to select $5$ of the $25$ students to go to the park and $binom{24}{4}$ ways for the teacher to select four other students to go the park with a particular student. Hence, each student is in $binom{24}{4}$ of the $binom{25}{5}$ possible groups the teacher can take to the park.
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
answered Dec 24 '18 at 1:34
N. F. TaussigN. F. Taussig
44.4k93357
44.4k93357
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
add a comment |
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
$begingroup$
Oh, Thank you so much for the effort. Understood your point. Realised my mistake. Thank you again.
$endgroup$
– jayant98
Dec 24 '18 at 1:39
add a comment |
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$begingroup$
The number of visit is simply $binom{25}{5}$. Since by symmetry every student goes the same number, each student goes 1/5 of total number of visits.
$endgroup$
– ablmf
Dec 24 '18 at 0:46
$begingroup$
@ablmf But, that's the problem I am not able to visualize the symmetry.
$endgroup$
– jayant98
Dec 24 '18 at 0:49
$begingroup$
A given student must visit the park with four of the other twenty-four students.
$endgroup$
– N. F. Taussig
Dec 24 '18 at 1:03
$begingroup$
@Taussig I am sorry. But, I am not able to understand it still. Maybe it is due to language problem (I have started learning English from about 10 months). If you may, please present an example. It may clear my thought. Thanks for your effort.
$endgroup$
– jayant98
Dec 24 '18 at 1:09
1
$begingroup$
By definition $^{25}C_5$ means the number of ways of choosing 5 from 25 which is what the question is asking. The symmetry in this case works because there is nothing to distinguish one student from another so the requirements and number of ways each student can be matched with 4 other students is identical for all 25. Symmetry would be broken if say, of the 5 students at least one has to be a girl. This would eliminate all boy groups of 5 and hence reduce the number of visits by the boys compared to the girls.
$endgroup$
– Phil H
Dec 24 '18 at 1:39