Combinatorial distribution [closed]
$begingroup$
Need some help with this exercise:
There are $300$ students in a course. Each student can get a grade from $0-100$.
How many ways can you divide the scores to get an average of $60$?
I had struggles to isolate the scenario of the exact required average.
After that I know how to make the distribution.
combinatorics
asked Dec 23 '18 at 12:48
IgorIgor
265
$endgroup$
closed as off-topic by Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500 Dec 31 '18 at 11:13
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Need some help with this exercise:
There are $300$ students in a course. Each student can get a grade from $0-100$.
How many ways can you divide the scores to get an average of $60$?
I had struggles to isolate the scenario of the exact required average.
After that I know how to make the distribution.
combinatorics
asked Dec 23 '18 at 12:48
IgorIgor
265
$endgroup$
closed as off-topic by Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500 Dec 31 '18 at 11:13
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
$endgroup$
– mrtaurho
Dec 23 '18 at 12:52
$begingroup$
Thanks you for the remark, I've edited it :)
$endgroup$
– Igor
Dec 23 '18 at 12:56
add a comment |
$begingroup$
Need some help with this exercise:
There are $300$ students in a course. Each student can get a grade from $0-100$.
How many ways can you divide the scores to get an average of $60$?
I had struggles to isolate the scenario of the exact required average.
After that I know how to make the distribution.
combinatorics
asked Dec 23 '18 at 12:48
IgorIgor
265
$endgroup$
Need some help with this exercise:
There are $300$ students in a course. Each student can get a grade from $0-100$.
How many ways can you divide the scores to get an average of $60$?
I had struggles to isolate the scenario of the exact required average.
After that I know how to make the distribution.
combinatorics
combinatorics
asked Dec 23 '18 at 12:48
IgorIgor
265
asked Dec 23 '18 at 12:48
IgorIgor
265
edited Dec 23 '18 at 18:15
Igor
asked Dec 23 '18 at 12:48
IgorIgor
265
asked Dec 23 '18 at 12:48
IgorIgor
265
asked Dec 23 '18 at 12:48
IgorIgor
265
265
closed as off-topic by Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500 Dec 31 '18 at 11:13
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500 Dec 31 '18 at 11:13
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, A. Pongrácz, Holo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
$endgroup$
– mrtaurho
Dec 23 '18 at 12:52
$begingroup$
Thanks you for the remark, I've edited it :)
$endgroup$
– Igor
Dec 23 '18 at 12:56
add a comment |
$begingroup$
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
$endgroup$
– mrtaurho
Dec 23 '18 at 12:52
$begingroup$
Thanks you for the remark, I've edited it :)
$endgroup$
– Igor
Dec 23 '18 at 12:56
$begingroup$
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
$endgroup$
– mrtaurho
Dec 23 '18 at 12:52
$begingroup$
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
$endgroup$
– mrtaurho
Dec 23 '18 at 12:52
$begingroup$
Thanks you for the remark, I've edited it :)
$endgroup$
– Igor
Dec 23 '18 at 12:56
$begingroup$
Thanks you for the remark, I've edited it :)
$endgroup$
– Igor
Dec 23 '18 at 12:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First of all note that average of 60 implies a total score (sum of score of all students) $= 60cdot300=18000$.
Also, let $x_i$ be the score of $i^{th}$ student. Then, $0leq x_ileq100$. I assume that you can only can integral grades.
Note that what you need is $x_1+x_2+cdots+x_{300}=18000$
So, using generating functions, total possible ways$=$
$$text{Coeff. of }z^{18000}text{ in }(z^0+z^1+z^2+cdots+z^{100})^{300}$$
$$=(frac{z^{101}-1}{z-1})^{300}$$
$$=(z^{101}-1)^{300}cdot(z-1)^{-300}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{300+(18000-101i)-1}{18000-101i}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{18299-101i}{18000-101i}$$
Note that I took $ileq 178$ because $18000-101igeq0implies ileq178.2$. And I'll be really honest. I don't know if there's any way to solve that summation.
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
$endgroup$
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
1
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
1
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
|
show 5 more comments
$begingroup$
Suppose there are $n$ students in a given course, and on a particular exam each student can receive an integer grade from $0$ to $100$. If the class average on the exam is exactly $60$, then how many different grade distributions are there for the students in the course?
What we are really doing is counting the solutions to a restricted Diophantine equation:
$$
s_1+s_2+cdots+s_n=ncdot60 quad text{with all } s_iin[0,100]
$$
In the case $n=1$, there is clearly only one solution. In the case $n=2$, we will have $81$ solutions, and for $ngeq 3$ it becomes clear that there are difficult counting problems involved that yield some rather large numbers.
Thankfully, this problem has been well studied; what we are really doing is counting lattice points in some high dimensional polytope, which can be accomplished with Ehrhart polynomials. In particular, this post has an answer that demonstrates how you may go about computing the answer you seek!
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
The correct generating function is already given by Ankit Kumar.
So to find the exact solution, a few lines of Mathematica code is enough:
g = (z^101 - 1)/(z - 1);
f = g^300;
SeriesCoefficient[f, {z, 0, 18000}]
283842333432402321353024303661058637351420936084067738626583506213864447957870755539611750086168273196143155816656392895618438969536797685183460631860648381858482057644617263985079629258986110897789595623952660661105103107276955285902613205328147901598400715242942785403171606118195123592165718190326762039521274137278366225940884773691094689005096899697160884662108869409844236284336801805326795531698837921700003716989423475547841836096863072991222376259898318028580853239559456607248039447660955555656440250262052607092640488420986795396794284727379957951608980168883759138931179704247147
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all note that average of 60 implies a total score (sum of score of all students) $= 60cdot300=18000$.
Also, let $x_i$ be the score of $i^{th}$ student. Then, $0leq x_ileq100$. I assume that you can only can integral grades.
Note that what you need is $x_1+x_2+cdots+x_{300}=18000$
So, using generating functions, total possible ways$=$
$$text{Coeff. of }z^{18000}text{ in }(z^0+z^1+z^2+cdots+z^{100})^{300}$$
$$=(frac{z^{101}-1}{z-1})^{300}$$
$$=(z^{101}-1)^{300}cdot(z-1)^{-300}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{300+(18000-101i)-1}{18000-101i}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{18299-101i}{18000-101i}$$
Note that I took $ileq 178$ because $18000-101igeq0implies ileq178.2$. And I'll be really honest. I don't know if there's any way to solve that summation.
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
$endgroup$
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
1
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
1
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
|
show 5 more comments
$begingroup$
First of all note that average of 60 implies a total score (sum of score of all students) $= 60cdot300=18000$.
Also, let $x_i$ be the score of $i^{th}$ student. Then, $0leq x_ileq100$. I assume that you can only can integral grades.
Note that what you need is $x_1+x_2+cdots+x_{300}=18000$
So, using generating functions, total possible ways$=$
$$text{Coeff. of }z^{18000}text{ in }(z^0+z^1+z^2+cdots+z^{100})^{300}$$
$$=(frac{z^{101}-1}{z-1})^{300}$$
$$=(z^{101}-1)^{300}cdot(z-1)^{-300}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{300+(18000-101i)-1}{18000-101i}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{18299-101i}{18000-101i}$$
Note that I took $ileq 178$ because $18000-101igeq0implies ileq178.2$. And I'll be really honest. I don't know if there's any way to solve that summation.
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
$endgroup$
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
1
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
1
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
|
show 5 more comments
$begingroup$
First of all note that average of 60 implies a total score (sum of score of all students) $= 60cdot300=18000$.
Also, let $x_i$ be the score of $i^{th}$ student. Then, $0leq x_ileq100$. I assume that you can only can integral grades.
Note that what you need is $x_1+x_2+cdots+x_{300}=18000$
So, using generating functions, total possible ways$=$
$$text{Coeff. of }z^{18000}text{ in }(z^0+z^1+z^2+cdots+z^{100})^{300}$$
$$=(frac{z^{101}-1}{z-1})^{300}$$
$$=(z^{101}-1)^{300}cdot(z-1)^{-300}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{300+(18000-101i)-1}{18000-101i}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{18299-101i}{18000-101i}$$
Note that I took $ileq 178$ because $18000-101igeq0implies ileq178.2$. And I'll be really honest. I don't know if there's any way to solve that summation.
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
$endgroup$
First of all note that average of 60 implies a total score (sum of score of all students) $= 60cdot300=18000$.
Also, let $x_i$ be the score of $i^{th}$ student. Then, $0leq x_ileq100$. I assume that you can only can integral grades.
Note that what you need is $x_1+x_2+cdots+x_{300}=18000$
So, using generating functions, total possible ways$=$
$$text{Coeff. of }z^{18000}text{ in }(z^0+z^1+z^2+cdots+z^{100})^{300}$$
$$=(frac{z^{101}-1}{z-1})^{300}$$
$$=(z^{101}-1)^{300}cdot(z-1)^{-300}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{300+(18000-101i)-1}{18000-101i}$$
$$=sum_{i=0}^{178}binom{300}{i}binom{18299-101i}{18000-101i}$$
Note that I took $ileq 178$ because $18000-101igeq0implies ileq178.2$. And I'll be really honest. I don't know if there's any way to solve that summation.
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
answered Dec 23 '18 at 13:53
Ankit KumarAnkit Kumar
1,540221
1,540221
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
1
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
1
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
|
show 5 more comments
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
1
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
1
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
I've came to something similar: $sumlimits_{j=0}^{300}binom{300+18000-100j-1}{18000-100j}binom{300}{j}(-1)^j$ Counting all the possibilities and limiting 100 for each "cell". After 180 I will just sum zeros.
$endgroup$
– Igor
Dec 23 '18 at 14:16
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
There's a slight mistake. $(-1)^{18000-101j}$ will come from the other term too. So you need not include that. Further, the formula for negative binomial is $binom{n+k-1}{k}$. You seem to have missed the -1. Also, it'll be 101j and not 100j because there're 101 terms in the geometric expression and not 100 ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 14:19
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
$begingroup$
-I don't understand why -1 is not necessary. - Yes, you are right :) - Why 101
$endgroup$
– Igor
Dec 23 '18 at 14:26
1
1
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
$begingroup$
This problem will be way easier for you to solve if you read about generating functions ;)
$endgroup$
– Ankit Kumar
Dec 23 '18 at 15:36
1
1
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
$begingroup$
This is the subject of the next lecture I will have this week :)
$endgroup$
– Igor
Dec 23 '18 at 17:32
|
show 5 more comments
$begingroup$
Suppose there are $n$ students in a given course, and on a particular exam each student can receive an integer grade from $0$ to $100$. If the class average on the exam is exactly $60$, then how many different grade distributions are there for the students in the course?
What we are really doing is counting the solutions to a restricted Diophantine equation:
$$
s_1+s_2+cdots+s_n=ncdot60 quad text{with all } s_iin[0,100]
$$
In the case $n=1$, there is clearly only one solution. In the case $n=2$, we will have $81$ solutions, and for $ngeq 3$ it becomes clear that there are difficult counting problems involved that yield some rather large numbers.
Thankfully, this problem has been well studied; what we are really doing is counting lattice points in some high dimensional polytope, which can be accomplished with Ehrhart polynomials. In particular, this post has an answer that demonstrates how you may go about computing the answer you seek!
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
Suppose there are $n$ students in a given course, and on a particular exam each student can receive an integer grade from $0$ to $100$. If the class average on the exam is exactly $60$, then how many different grade distributions are there for the students in the course?
What we are really doing is counting the solutions to a restricted Diophantine equation:
$$
s_1+s_2+cdots+s_n=ncdot60 quad text{with all } s_iin[0,100]
$$
In the case $n=1$, there is clearly only one solution. In the case $n=2$, we will have $81$ solutions, and for $ngeq 3$ it becomes clear that there are difficult counting problems involved that yield some rather large numbers.
Thankfully, this problem has been well studied; what we are really doing is counting lattice points in some high dimensional polytope, which can be accomplished with Ehrhart polynomials. In particular, this post has an answer that demonstrates how you may go about computing the answer you seek!
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
Suppose there are $n$ students in a given course, and on a particular exam each student can receive an integer grade from $0$ to $100$. If the class average on the exam is exactly $60$, then how many different grade distributions are there for the students in the course?
What we are really doing is counting the solutions to a restricted Diophantine equation:
$$
s_1+s_2+cdots+s_n=ncdot60 quad text{with all } s_iin[0,100]
$$
In the case $n=1$, there is clearly only one solution. In the case $n=2$, we will have $81$ solutions, and for $ngeq 3$ it becomes clear that there are difficult counting problems involved that yield some rather large numbers.
Thankfully, this problem has been well studied; what we are really doing is counting lattice points in some high dimensional polytope, which can be accomplished with Ehrhart polynomials. In particular, this post has an answer that demonstrates how you may go about computing the answer you seek!
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
$endgroup$
Suppose there are $n$ students in a given course, and on a particular exam each student can receive an integer grade from $0$ to $100$. If the class average on the exam is exactly $60$, then how many different grade distributions are there for the students in the course?
What we are really doing is counting the solutions to a restricted Diophantine equation:
$$
s_1+s_2+cdots+s_n=ncdot60 quad text{with all } s_iin[0,100]
$$
In the case $n=1$, there is clearly only one solution. In the case $n=2$, we will have $81$ solutions, and for $ngeq 3$ it becomes clear that there are difficult counting problems involved that yield some rather large numbers.
Thankfully, this problem has been well studied; what we are really doing is counting lattice points in some high dimensional polytope, which can be accomplished with Ehrhart polynomials. In particular, this post has an answer that demonstrates how you may go about computing the answer you seek!
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
edited Dec 23 '18 at 14:50
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 13:24
RandomNumberGuyRandomNumberGuy
662
662
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The correct generating function is already given by Ankit Kumar.
So to find the exact solution, a few lines of Mathematica code is enough:
g = (z^101 - 1)/(z - 1);
f = g^300;
SeriesCoefficient[f, {z, 0, 18000}]
283842333432402321353024303661058637351420936084067738626583506213864447957870755539611750086168273196143155816656392895618438969536797685183460631860648381858482057644617263985079629258986110897789595623952660661105103107276955285902613205328147901598400715242942785403171606118195123592165718190326762039521274137278366225940884773691094689005096899697160884662108869409844236284336801805326795531698837921700003716989423475547841836096863072991222376259898318028580853239559456607248039447660955555656440250262052607092640488420986795396794284727379957951608980168883759138931179704247147
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
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add a comment |
$begingroup$
The correct generating function is already given by Ankit Kumar.
So to find the exact solution, a few lines of Mathematica code is enough:
g = (z^101 - 1)/(z - 1);
f = g^300;
SeriesCoefficient[f, {z, 0, 18000}]
283842333432402321353024303661058637351420936084067738626583506213864447957870755539611750086168273196143155816656392895618438969536797685183460631860648381858482057644617263985079629258986110897789595623952660661105103107276955285902613205328147901598400715242942785403171606118195123592165718190326762039521274137278366225940884773691094689005096899697160884662108869409844236284336801805326795531698837921700003716989423475547841836096863072991222376259898318028580853239559456607248039447660955555656440250262052607092640488420986795396794284727379957951608980168883759138931179704247147
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
$endgroup$
add a comment |
$begingroup$
The correct generating function is already given by Ankit Kumar.
So to find the exact solution, a few lines of Mathematica code is enough:
g = (z^101 - 1)/(z - 1);
f = g^300;
SeriesCoefficient[f, {z, 0, 18000}]
283842333432402321353024303661058637351420936084067738626583506213864447957870755539611750086168273196143155816656392895618438969536797685183460631860648381858482057644617263985079629258986110897789595623952660661105103107276955285902613205328147901598400715242942785403171606118195123592165718190326762039521274137278366225940884773691094689005096899697160884662108869409844236284336801805326795531698837921700003716989423475547841836096863072991222376259898318028580853239559456607248039447660955555656440250262052607092640488420986795396794284727379957951608980168883759138931179704247147
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
$endgroup$
The correct generating function is already given by Ankit Kumar.
So to find the exact solution, a few lines of Mathematica code is enough:
g = (z^101 - 1)/(z - 1);
f = g^300;
SeriesCoefficient[f, {z, 0, 18000}]
283842333432402321353024303661058637351420936084067738626583506213864447957870755539611750086168273196143155816656392895618438969536797685183460631860648381858482057644617263985079629258986110897789595623952660661105103107276955285902613205328147901598400715242942785403171606118195123592165718190326762039521274137278366225940884773691094689005096899697160884662108869409844236284336801805326795531698837921700003716989423475547841836096863072991222376259898318028580853239559456607248039447660955555656440250262052607092640488420986795396794284727379957951608980168883759138931179704247147
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
answered Dec 23 '18 at 22:19
ablmfablmf
2,58352452
2,58352452
add a comment |
add a comment |
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Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
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– mrtaurho
Dec 23 '18 at 12:52
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Thanks you for the remark, I've edited it :)
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– Igor
Dec 23 '18 at 12:56