In how many ways can you grade 5 students, so their average is 60? [closed]
$begingroup$
I've been struggling with this question:
In how many ways can you grade 5 students so their average is 60? Note that each student's grade is an integer, no greater than 100 and no lower than 0.
I know that no more than 3 students can get a 100, but I don't see a way to count all the possible permutations for dividing 300 points into 5 baskets, so each of them is no more than 100 points.
I'd be grateful if you could hint me the right direction for solving this problem.
combinatorics
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
$endgroup$
closed as off-topic by Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo Dec 26 '18 at 9:52
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." – Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
I've been struggling with this question:
In how many ways can you grade 5 students so their average is 60? Note that each student's grade is an integer, no greater than 100 and no lower than 0.
I know that no more than 3 students can get a 100, but I don't see a way to count all the possible permutations for dividing 300 points into 5 baskets, so each of them is no more than 100 points.
I'd be grateful if you could hint me the right direction for solving this problem.
combinatorics
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
$endgroup$
closed as off-topic by Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo Dec 26 '18 at 9:52
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." – Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What other constraints are there on the grades? I presume they need to be integers, but is this correct?
$endgroup$
– Cameron Buie
Dec 25 '18 at 15:40
$begingroup$
yes. i should have mentioned it.
$endgroup$
– Ido Doron
Dec 25 '18 at 15:41
4
$begingroup$
You can adapt the technique presented in this answer, but with your numbers and constraints.
$endgroup$
– Don Thousand
Dec 25 '18 at 15:42
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 15:42
$begingroup$
By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways
$endgroup$
– WaveX
Dec 25 '18 at 16:00
|
show 1 more comment
$begingroup$
I've been struggling with this question:
In how many ways can you grade 5 students so their average is 60? Note that each student's grade is an integer, no greater than 100 and no lower than 0.
I know that no more than 3 students can get a 100, but I don't see a way to count all the possible permutations for dividing 300 points into 5 baskets, so each of them is no more than 100 points.
I'd be grateful if you could hint me the right direction for solving this problem.
combinatorics
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
$endgroup$
I've been struggling with this question:
In how many ways can you grade 5 students so their average is 60? Note that each student's grade is an integer, no greater than 100 and no lower than 0.
I know that no more than 3 students can get a 100, but I don't see a way to count all the possible permutations for dividing 300 points into 5 baskets, so each of them is no more than 100 points.
I'd be grateful if you could hint me the right direction for solving this problem.
combinatorics
combinatorics
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
edited Dec 25 '18 at 15:52
Ido Doron
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
asked Dec 25 '18 at 15:39
Ido DoronIdo Doron
172
172
closed as off-topic by Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo Dec 26 '18 at 9:52
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." – Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo Dec 26 '18 at 9:52
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." – Shaun, Don Thousand, Eevee Trainer, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What other constraints are there on the grades? I presume they need to be integers, but is this correct?
$endgroup$
– Cameron Buie
Dec 25 '18 at 15:40
$begingroup$
yes. i should have mentioned it.
$endgroup$
– Ido Doron
Dec 25 '18 at 15:41
4
$begingroup$
You can adapt the technique presented in this answer, but with your numbers and constraints.
$endgroup$
– Don Thousand
Dec 25 '18 at 15:42
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 15:42
$begingroup$
By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways
$endgroup$
– WaveX
Dec 25 '18 at 16:00
|
show 1 more comment
$begingroup$
What other constraints are there on the grades? I presume they need to be integers, but is this correct?
$endgroup$
– Cameron Buie
Dec 25 '18 at 15:40
$begingroup$
yes. i should have mentioned it.
$endgroup$
– Ido Doron
Dec 25 '18 at 15:41
4
$begingroup$
You can adapt the technique presented in this answer, but with your numbers and constraints.
$endgroup$
– Don Thousand
Dec 25 '18 at 15:42
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 15:42
$begingroup$
By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways
$endgroup$
– WaveX
Dec 25 '18 at 16:00
$begingroup$
What other constraints are there on the grades? I presume they need to be integers, but is this correct?
$endgroup$
– Cameron Buie
Dec 25 '18 at 15:40
$begingroup$
What other constraints are there on the grades? I presume they need to be integers, but is this correct?
$endgroup$
– Cameron Buie
Dec 25 '18 at 15:40
$begingroup$
yes. i should have mentioned it.
$endgroup$
– Ido Doron
Dec 25 '18 at 15:41
$begingroup$
yes. i should have mentioned it.
$endgroup$
– Ido Doron
Dec 25 '18 at 15:41
4
4
$begingroup$
You can adapt the technique presented in this answer, but with your numbers and constraints.
$endgroup$
– Don Thousand
Dec 25 '18 at 15:42
$begingroup$
You can adapt the technique presented in this answer, but with your numbers and constraints.
$endgroup$
– Don Thousand
Dec 25 '18 at 15:42
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 15:42
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 15:42
$begingroup$
By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways
$endgroup$
– WaveX
Dec 25 '18 at 16:00
$begingroup$
By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways
$endgroup$
– WaveX
Dec 25 '18 at 16:00
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
We can also do it by generating function.
The generating function for the same is $(1+x+x^2+.. + x^{100})^{5}$ and what we want is the coefficient of $x^{300}$.
$(1+x+x^2+.. + x^{100})^{5} = (1-x^{101})^{5}.frac{1}{(1-x)^{5}}$
$(1-x^{101})^{5}$ can be expressed $(1-{5choose1}x^{101} + {5choose2}x^{202}-{5choose3}x^{303}..$
$$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+5-1)choose(5-1)}x^n$$
$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+4)choose(4)}x^n$
Multiplying these two expressions, you are looking to have n+4-r where r is the reducing number from the first expression. $r_i = 0, 101, 202$ for $r i = 0,1,2$ respectively
Thus coefficients are products of $(-1)^i({5choose i}{(300+4-r_i)choose 4}) x^{300}$
you get ${(300+4)choose(4)}$ for the first term
you get the next one $- {5choose1}{(300+4-101)choose(4)}$
and the next one ${5choose2}{(300+4-202)choose(4)}$
Add these products such as below
$ {304choose4}- {5choose1}{203choose4}+
{5choose2}{102choose(4)}=47952376 $
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
add a comment |
$begingroup$
I just made a spreadsheet. Start in row $101$, leaving all the ones above blank, and enter the numbers from $0$ through $300$ in column A. That will be the sum of grades. Now in column B enter $1$ in all the rows with numbers $0$ through $100$ in column A, which shows there is one way to give the first student any number of points from $0$ through $100$. Now in column C in line with $0$ points enter =SUM(Left up 100:Left), where you substitute the cell references for the directions, which says the number of ways to get a sum of $0$ for the first two students is the sum of the $101$ lines ending in this one. Copy down and right for $300$ points and $5$ students. I find the total to be $45235674$
The generating function approach is that the function for one student is $1+x+x^2+ldots x^{100}=frac {1-x^{101}}{1-x}$ so the function for five students is $$left(frac {1-x^{101}}{1-x}right)^5$$
and you want the coefficient of $x^{300}$ of this. I don't have an easy way to evaluate that, but I think Mathematica can do so.
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
$begingroup$
CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]
$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
$begingroup$
@David G Stork,Could you let me know how it evaluates to?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:12
$begingroup$
@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
$endgroup$
– David G. Stork
Dec 25 '18 at 18:46
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can also do it by generating function.
The generating function for the same is $(1+x+x^2+.. + x^{100})^{5}$ and what we want is the coefficient of $x^{300}$.
$(1+x+x^2+.. + x^{100})^{5} = (1-x^{101})^{5}.frac{1}{(1-x)^{5}}$
$(1-x^{101})^{5}$ can be expressed $(1-{5choose1}x^{101} + {5choose2}x^{202}-{5choose3}x^{303}..$
$$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+5-1)choose(5-1)}x^n$$
$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+4)choose(4)}x^n$
Multiplying these two expressions, you are looking to have n+4-r where r is the reducing number from the first expression. $r_i = 0, 101, 202$ for $r i = 0,1,2$ respectively
Thus coefficients are products of $(-1)^i({5choose i}{(300+4-r_i)choose 4}) x^{300}$
you get ${(300+4)choose(4)}$ for the first term
you get the next one $- {5choose1}{(300+4-101)choose(4)}$
and the next one ${5choose2}{(300+4-202)choose(4)}$
Add these products such as below
$ {304choose4}- {5choose1}{203choose4}+
{5choose2}{102choose(4)}=47952376 $
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
add a comment |
$begingroup$
We can also do it by generating function.
The generating function for the same is $(1+x+x^2+.. + x^{100})^{5}$ and what we want is the coefficient of $x^{300}$.
$(1+x+x^2+.. + x^{100})^{5} = (1-x^{101})^{5}.frac{1}{(1-x)^{5}}$
$(1-x^{101})^{5}$ can be expressed $(1-{5choose1}x^{101} + {5choose2}x^{202}-{5choose3}x^{303}..$
$$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+5-1)choose(5-1)}x^n$$
$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+4)choose(4)}x^n$
Multiplying these two expressions, you are looking to have n+4-r where r is the reducing number from the first expression. $r_i = 0, 101, 202$ for $r i = 0,1,2$ respectively
Thus coefficients are products of $(-1)^i({5choose i}{(300+4-r_i)choose 4}) x^{300}$
you get ${(300+4)choose(4)}$ for the first term
you get the next one $- {5choose1}{(300+4-101)choose(4)}$
and the next one ${5choose2}{(300+4-202)choose(4)}$
Add these products such as below
$ {304choose4}- {5choose1}{203choose4}+
{5choose2}{102choose(4)}=47952376 $
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
add a comment |
$begingroup$
We can also do it by generating function.
The generating function for the same is $(1+x+x^2+.. + x^{100})^{5}$ and what we want is the coefficient of $x^{300}$.
$(1+x+x^2+.. + x^{100})^{5} = (1-x^{101})^{5}.frac{1}{(1-x)^{5}}$
$(1-x^{101})^{5}$ can be expressed $(1-{5choose1}x^{101} + {5choose2}x^{202}-{5choose3}x^{303}..$
$$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+5-1)choose(5-1)}x^n$$
$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+4)choose(4)}x^n$
Multiplying these two expressions, you are looking to have n+4-r where r is the reducing number from the first expression. $r_i = 0, 101, 202$ for $r i = 0,1,2$ respectively
Thus coefficients are products of $(-1)^i({5choose i}{(300+4-r_i)choose 4}) x^{300}$
you get ${(300+4)choose(4)}$ for the first term
you get the next one $- {5choose1}{(300+4-101)choose(4)}$
and the next one ${5choose2}{(300+4-202)choose(4)}$
Add these products such as below
$ {304choose4}- {5choose1}{203choose4}+
{5choose2}{102choose(4)}=47952376 $
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
We can also do it by generating function.
The generating function for the same is $(1+x+x^2+.. + x^{100})^{5}$ and what we want is the coefficient of $x^{300}$.
$(1+x+x^2+.. + x^{100})^{5} = (1-x^{101})^{5}.frac{1}{(1-x)^{5}}$
$(1-x^{101})^{5}$ can be expressed $(1-{5choose1}x^{101} + {5choose2}x^{202}-{5choose3}x^{303}..$
$$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+5-1)choose(5-1)}x^n$$
$frac{1}{(1-x)^{5}} = sum_{0}^{infty} {(n+4)choose(4)}x^n$
Multiplying these two expressions, you are looking to have n+4-r where r is the reducing number from the first expression. $r_i = 0, 101, 202$ for $r i = 0,1,2$ respectively
Thus coefficients are products of $(-1)^i({5choose i}{(300+4-r_i)choose 4}) x^{300}$
you get ${(300+4)choose(4)}$ for the first term
you get the next one $- {5choose1}{(300+4-101)choose(4)}$
and the next one ${5choose2}{(300+4-202)choose(4)}$
Add these products such as below
$ {304choose4}- {5choose1}{203choose4}+
{5choose2}{102choose(4)}=47952376 $
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
edited Dec 26 '18 at 3:45
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
answered Dec 25 '18 at 17:10
Satish RamanathanSatish Ramanathan
9,92531323
9,92531323
add a comment |
add a comment |
$begingroup$
I just made a spreadsheet. Start in row $101$, leaving all the ones above blank, and enter the numbers from $0$ through $300$ in column A. That will be the sum of grades. Now in column B enter $1$ in all the rows with numbers $0$ through $100$ in column A, which shows there is one way to give the first student any number of points from $0$ through $100$. Now in column C in line with $0$ points enter =SUM(Left up 100:Left), where you substitute the cell references for the directions, which says the number of ways to get a sum of $0$ for the first two students is the sum of the $101$ lines ending in this one. Copy down and right for $300$ points and $5$ students. I find the total to be $45235674$
The generating function approach is that the function for one student is $1+x+x^2+ldots x^{100}=frac {1-x^{101}}{1-x}$ so the function for five students is $$left(frac {1-x^{101}}{1-x}right)^5$$
and you want the coefficient of $x^{300}$ of this. I don't have an easy way to evaluate that, but I think Mathematica can do so.
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
$begingroup$
CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]
$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
$begingroup$
@David G Stork,Could you let me know how it evaluates to?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:12
$begingroup$
@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
$endgroup$
– David G. Stork
Dec 25 '18 at 18:46
add a comment |
$begingroup$
I just made a spreadsheet. Start in row $101$, leaving all the ones above blank, and enter the numbers from $0$ through $300$ in column A. That will be the sum of grades. Now in column B enter $1$ in all the rows with numbers $0$ through $100$ in column A, which shows there is one way to give the first student any number of points from $0$ through $100$. Now in column C in line with $0$ points enter =SUM(Left up 100:Left), where you substitute the cell references for the directions, which says the number of ways to get a sum of $0$ for the first two students is the sum of the $101$ lines ending in this one. Copy down and right for $300$ points and $5$ students. I find the total to be $45235674$
The generating function approach is that the function for one student is $1+x+x^2+ldots x^{100}=frac {1-x^{101}}{1-x}$ so the function for five students is $$left(frac {1-x^{101}}{1-x}right)^5$$
and you want the coefficient of $x^{300}$ of this. I don't have an easy way to evaluate that, but I think Mathematica can do so.
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
$begingroup$
CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]
$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
$begingroup$
@David G Stork,Could you let me know how it evaluates to?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:12
$begingroup$
@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
$endgroup$
– David G. Stork
Dec 25 '18 at 18:46
add a comment |
$begingroup$
I just made a spreadsheet. Start in row $101$, leaving all the ones above blank, and enter the numbers from $0$ through $300$ in column A. That will be the sum of grades. Now in column B enter $1$ in all the rows with numbers $0$ through $100$ in column A, which shows there is one way to give the first student any number of points from $0$ through $100$. Now in column C in line with $0$ points enter =SUM(Left up 100:Left), where you substitute the cell references for the directions, which says the number of ways to get a sum of $0$ for the first two students is the sum of the $101$ lines ending in this one. Copy down and right for $300$ points and $5$ students. I find the total to be $45235674$
The generating function approach is that the function for one student is $1+x+x^2+ldots x^{100}=frac {1-x^{101}}{1-x}$ so the function for five students is $$left(frac {1-x^{101}}{1-x}right)^5$$
and you want the coefficient of $x^{300}$ of this. I don't have an easy way to evaluate that, but I think Mathematica can do so.
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
I just made a spreadsheet. Start in row $101$, leaving all the ones above blank, and enter the numbers from $0$ through $300$ in column A. That will be the sum of grades. Now in column B enter $1$ in all the rows with numbers $0$ through $100$ in column A, which shows there is one way to give the first student any number of points from $0$ through $100$. Now in column C in line with $0$ points enter =SUM(Left up 100:Left), where you substitute the cell references for the directions, which says the number of ways to get a sum of $0$ for the first two students is the sum of the $101$ lines ending in this one. Copy down and right for $300$ points and $5$ students. I find the total to be $45235674$
The generating function approach is that the function for one student is $1+x+x^2+ldots x^{100}=frac {1-x^{101}}{1-x}$ so the function for five students is $$left(frac {1-x^{101}}{1-x}right)^5$$
and you want the coefficient of $x^{300}$ of this. I don't have an easy way to evaluate that, but I think Mathematica can do so.
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
edited Dec 26 '18 at 1:10
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
answered Dec 25 '18 at 16:16
Ross MillikanRoss Millikan
296k23198371
296k23198371
$begingroup$
CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]
$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
$begingroup$
@David G Stork,Could you let me know how it evaluates to?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:12
$begingroup$
@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
$endgroup$
– David G. Stork
Dec 25 '18 at 18:46
add a comment |
$begingroup$
CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]
$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
$begingroup$
@David G Stork,Could you let me know how it evaluates to?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:12
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@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
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– David G. Stork
Dec 25 '18 at 18:46
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CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
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CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]$endgroup$
– David G. Stork
Dec 25 '18 at 16:55
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@David G Stork,Could you let me know how it evaluates to?
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– Satish Ramanathan
Dec 25 '18 at 17:12
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@David G Stork,Could you let me know how it evaluates to?
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– Satish Ramanathan
Dec 25 '18 at 17:12
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@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
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– David G. Stork
Dec 25 '18 at 18:46
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@SatishRamanathan: 48462325, but I realize that cannot be correct, because the partitions are ordered, but the students are not: so it counts {40,50,60,70,80} as the same as {80,70,60,50,40}. One must reduce by certain numbers of permutations.
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– David G. Stork
Dec 25 '18 at 18:46
add a comment |
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What other constraints are there on the grades? I presume they need to be integers, but is this correct?
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– Cameron Buie
Dec 25 '18 at 15:40
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yes. i should have mentioned it.
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– Ido Doron
Dec 25 '18 at 15:41
4
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You can adapt the technique presented in this answer, but with your numbers and constraints.
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– Don Thousand
Dec 25 '18 at 15:42
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You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
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– Shaun
Dec 25 '18 at 15:42
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By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways
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– WaveX
Dec 25 '18 at 16:00