In how many ways can you grade 5 students, so their average is 60? [closed]












-1












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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.










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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


















-1












$begingroup$


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.










share|cite|improve this question











$endgroup$



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
















-1












-1








-1





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 15:52







Ido Doron

















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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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 $






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $






        share|cite|improve this answer











        $endgroup$



        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 $







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 26 '18 at 3:45

























        answered Dec 25 '18 at 17:10









        Satish RamanathanSatish Ramanathan

        10k31323




        10k31323























            0












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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


















            0












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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
















            0












            0








            0





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 26 '18 at 1:10

























            answered Dec 25 '18 at 16:16









            Ross MillikanRoss Millikan

            297k23198371




            297k23198371












            • $begingroup$
              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




















            • $begingroup$
              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


















            $begingroup$
            CoefficientList[((1 - x^101)/(1 - x))^5, x][[300]]
            $endgroup$
            – David G. Stork
            Dec 25 '18 at 16:55




            $begingroup$
            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




            $begingroup$
            @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






            $begingroup$
            @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





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