How many integer solutions are there for $x_1+x_2+x_3+x_4+x_5=75$, with $1 leq i leq 5, x_1=x_5$?












0












$begingroup$


How many integer solutions are there for $x_1+x_2+x_3+x_4+x_5=75$, with $1leq i leq 5,$ and $ x_1=x_5$ ?



I think I understand that if $x_1=x_5$ wasn't part of it, then it would be $binom{n-1}{k-1}.$



But if someone could explain how I deal with the $x_1=x_5$ that would be great!






combinatorics combinations







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$endgroup$












  • $begingroup$
    In your "understanding," what would $n$ and $k$ be for this case? Does your formula work for a simple small case, such as $x_1 + x_2 + x_3 = 5$? And why not simply ask your question: $2 x_1 + x_2 + x_3 + x_4 = 75$?
    $endgroup$
    – David G. Stork
    Jan 22 at 19:30








  • 1




    $begingroup$
    Think of distributing 75 cookies among five people. Except among those five people there are two brothers who need to receive the same number of cookies! Feel free to ask for any clarifications and I shall help!
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:36










  • $begingroup$
    To my understanding n = 5 and k = 3.
    $endgroup$
    – EveresttML
    Jan 22 at 19:47
















0












$begingroup$


How many integer solutions are there for $x_1+x_2+x_3+x_4+x_5=75$, with $1leq i leq 5,$ and $ x_1=x_5$ ?



I think I understand that if $x_1=x_5$ wasn't part of it, then it would be $binom{n-1}{k-1}.$



But if someone could explain how I deal with the $x_1=x_5$ that would be great!






combinatorics combinations







share|cite|improve this question











$endgroup$












  • $begingroup$
    In your "understanding," what would $n$ and $k$ be for this case? Does your formula work for a simple small case, such as $x_1 + x_2 + x_3 = 5$? And why not simply ask your question: $2 x_1 + x_2 + x_3 + x_4 = 75$?
    $endgroup$
    – David G. Stork
    Jan 22 at 19:30








  • 1




    $begingroup$
    Think of distributing 75 cookies among five people. Except among those five people there are two brothers who need to receive the same number of cookies! Feel free to ask for any clarifications and I shall help!
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:36










  • $begingroup$
    To my understanding n = 5 and k = 3.
    $endgroup$
    – EveresttML
    Jan 22 at 19:47














0












0








0





$begingroup$


How many integer solutions are there for $x_1+x_2+x_3+x_4+x_5=75$, with $1leq i leq 5,$ and $ x_1=x_5$ ?



I think I understand that if $x_1=x_5$ wasn't part of it, then it would be $binom{n-1}{k-1}.$



But if someone could explain how I deal with the $x_1=x_5$ that would be great!






combinatorics combinations







share|cite|improve this question











$endgroup$




How many integer solutions are there for $x_1+x_2+x_3+x_4+x_5=75$, with $1leq i leq 5,$ and $ x_1=x_5$ ?



I think I understand that if $x_1=x_5$ wasn't part of it, then it would be $binom{n-1}{k-1}.$



But if someone could explain how I deal with the $x_1=x_5$ that would be great!






combinatorics combinations




combinatorics combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:33









jordan_glen

1




1










asked Jan 22 at 19:24









EveresttMLEveresttML

11




11












  • $begingroup$
    In your "understanding," what would $n$ and $k$ be for this case? Does your formula work for a simple small case, such as $x_1 + x_2 + x_3 = 5$? And why not simply ask your question: $2 x_1 + x_2 + x_3 + x_4 = 75$?
    $endgroup$
    – David G. Stork
    Jan 22 at 19:30








  • 1




    $begingroup$
    Think of distributing 75 cookies among five people. Except among those five people there are two brothers who need to receive the same number of cookies! Feel free to ask for any clarifications and I shall help!
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:36










  • $begingroup$
    To my understanding n = 5 and k = 3.
    $endgroup$
    – EveresttML
    Jan 22 at 19:47


















  • $begingroup$
    In your "understanding," what would $n$ and $k$ be for this case? Does your formula work for a simple small case, such as $x_1 + x_2 + x_3 = 5$? And why not simply ask your question: $2 x_1 + x_2 + x_3 + x_4 = 75$?
    $endgroup$
    – David G. Stork
    Jan 22 at 19:30








  • 1




    $begingroup$
    Think of distributing 75 cookies among five people. Except among those five people there are two brothers who need to receive the same number of cookies! Feel free to ask for any clarifications and I shall help!
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:36










  • $begingroup$
    To my understanding n = 5 and k = 3.
    $endgroup$
    – EveresttML
    Jan 22 at 19:47
















$begingroup$
In your "understanding," what would $n$ and $k$ be for this case? Does your formula work for a simple small case, such as $x_1 + x_2 + x_3 = 5$? And why not simply ask your question: $2 x_1 + x_2 + x_3 + x_4 = 75$?
$endgroup$
– David G. Stork
Jan 22 at 19:30






$begingroup$
In your "understanding," what would $n$ and $k$ be for this case? Does your formula work for a simple small case, such as $x_1 + x_2 + x_3 = 5$? And why not simply ask your question: $2 x_1 + x_2 + x_3 + x_4 = 75$?
$endgroup$
– David G. Stork
Jan 22 at 19:30






1




1




$begingroup$
Think of distributing 75 cookies among five people. Except among those five people there are two brothers who need to receive the same number of cookies! Feel free to ask for any clarifications and I shall help!
$endgroup$
– The Jade Emperor
Jan 22 at 19:36




$begingroup$
Think of distributing 75 cookies among five people. Except among those five people there are two brothers who need to receive the same number of cookies! Feel free to ask for any clarifications and I shall help!
$endgroup$
– The Jade Emperor
Jan 22 at 19:36












$begingroup$
To my understanding n = 5 and k = 3.
$endgroup$
– EveresttML
Jan 22 at 19:47




$begingroup$
To my understanding n = 5 and k = 3.
$endgroup$
– EveresttML
Jan 22 at 19:47










3 Answers
3






active

oldest

votes


















0












$begingroup$

Remember if $x_1=x_5$ this expression can be rewritten as $2x_1+x_2+x_3+x_4=75$



This implies $3leq x_2+x_3+x_4leq73$ and that sum is odd. Can you find how many ways this is true






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would look better in a comment.. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:37



















0












$begingroup$

Hint: Consider calculating how many positive integer solutions there are to $$2x_1 + x_2 + x_3 + x_4= 75;?$$






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    This would look better in a comment.. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:37



















0












$begingroup$

I haven't done combinatorics in a long time so I might not give you the correct answer but here are some thoughts:



First off, perhaps you should clarify/consider if zeros are allowed. Let's suppose this is so. Think about how your formula can be justified combinatorially: using balls and walls. You can read that here: https://math.dartmouth.edu/archive/m68f09/public_html/m68l02.pdf



I would think of a simpler case. writing 5 as a sum of 4 parts c1, c2, c3 and c4. My idea would then be that we would have 3 cases: c1=c4=0,1 or 2 (3 won't do it). Also notice that I'm using a multiple of 5 for a reason. I guess you would have to take the floor or ceiling function of 75/2.



We're gonna have to use 3 walls (k-1 in your formula). Case 1 looks like this:
$underset{c1}{}|underset{c2}{text{some balls}}|underset{c3}{text{some balls}}|underset{c4}{}$. Of course, under "some balls" we know we have 5 of them and
c1 and c4 are chosen to have none. The problem then becomes: in how many ways can we write 5 has a sum of 2 nonnegative integers? Here, n=5 and k=1.



Case 2 has c1=c4=1 so we have the question: in how many ways can we write 5-2=3 as a sum of 2 nonnegative integers (remaining balls)? Run through all the cases and add them up.



Does this makes sense? Maybe there's a way easier way to solve this but hey, its an idea. What I'm saying is that my strategy would be to consider the actual proof and see what modifications need to be made






share|cite|improve this answer











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

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    active

    oldest

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    0












    $begingroup$

    Remember if $x_1=x_5$ this expression can be rewritten as $2x_1+x_2+x_3+x_4=75$



    This implies $3leq x_2+x_3+x_4leq73$ and that sum is odd. Can you find how many ways this is true






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37
















    0












    $begingroup$

    Remember if $x_1=x_5$ this expression can be rewritten as $2x_1+x_2+x_3+x_4=75$



    This implies $3leq x_2+x_3+x_4leq73$ and that sum is odd. Can you find how many ways this is true






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37














    0












    0








    0





    $begingroup$

    Remember if $x_1=x_5$ this expression can be rewritten as $2x_1+x_2+x_3+x_4=75$



    This implies $3leq x_2+x_3+x_4leq73$ and that sum is odd. Can you find how many ways this is true






    share|cite|improve this answer









    $endgroup$



    Remember if $x_1=x_5$ this expression can be rewritten as $2x_1+x_2+x_3+x_4=75$



    This implies $3leq x_2+x_3+x_4leq73$ and that sum is odd. Can you find how many ways this is true







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 22 at 19:36









    Rhys HughesRhys Hughes

    6,9441530




    6,9441530












    • $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37


















    • $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37
















    $begingroup$
    This would look better in a comment.. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:37




    $begingroup$
    This would look better in a comment.. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:37











    0












    $begingroup$

    Hint: Consider calculating how many positive integer solutions there are to $$2x_1 + x_2 + x_3 + x_4= 75;?$$






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37
















    0












    $begingroup$

    Hint: Consider calculating how many positive integer solutions there are to $$2x_1 + x_2 + x_3 + x_4= 75;?$$






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37














    0












    0








    0





    $begingroup$

    Hint: Consider calculating how many positive integer solutions there are to $$2x_1 + x_2 + x_3 + x_4= 75;?$$






    share|cite|improve this answer











    $endgroup$



    Hint: Consider calculating how many positive integer solutions there are to $$2x_1 + x_2 + x_3 + x_4= 75;?$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 22 at 19:38

























    answered Jan 22 at 19:36









    jordan_glenjordan_glen

    1




    1








    • 2




      $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37














    • 2




      $begingroup$
      This would look better in a comment.. :)
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:37








    2




    2




    $begingroup$
    This would look better in a comment.. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:37




    $begingroup$
    This would look better in a comment.. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:37











    0












    $begingroup$

    I haven't done combinatorics in a long time so I might not give you the correct answer but here are some thoughts:



    First off, perhaps you should clarify/consider if zeros are allowed. Let's suppose this is so. Think about how your formula can be justified combinatorially: using balls and walls. You can read that here: https://math.dartmouth.edu/archive/m68f09/public_html/m68l02.pdf



    I would think of a simpler case. writing 5 as a sum of 4 parts c1, c2, c3 and c4. My idea would then be that we would have 3 cases: c1=c4=0,1 or 2 (3 won't do it). Also notice that I'm using a multiple of 5 for a reason. I guess you would have to take the floor or ceiling function of 75/2.



    We're gonna have to use 3 walls (k-1 in your formula). Case 1 looks like this:
    $underset{c1}{}|underset{c2}{text{some balls}}|underset{c3}{text{some balls}}|underset{c4}{}$. Of course, under "some balls" we know we have 5 of them and
    c1 and c4 are chosen to have none. The problem then becomes: in how many ways can we write 5 has a sum of 2 nonnegative integers? Here, n=5 and k=1.



    Case 2 has c1=c4=1 so we have the question: in how many ways can we write 5-2=3 as a sum of 2 nonnegative integers (remaining balls)? Run through all the cases and add them up.



    Does this makes sense? Maybe there's a way easier way to solve this but hey, its an idea. What I'm saying is that my strategy would be to consider the actual proof and see what modifications need to be made






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      I haven't done combinatorics in a long time so I might not give you the correct answer but here are some thoughts:



      First off, perhaps you should clarify/consider if zeros are allowed. Let's suppose this is so. Think about how your formula can be justified combinatorially: using balls and walls. You can read that here: https://math.dartmouth.edu/archive/m68f09/public_html/m68l02.pdf



      I would think of a simpler case. writing 5 as a sum of 4 parts c1, c2, c3 and c4. My idea would then be that we would have 3 cases: c1=c4=0,1 or 2 (3 won't do it). Also notice that I'm using a multiple of 5 for a reason. I guess you would have to take the floor or ceiling function of 75/2.



      We're gonna have to use 3 walls (k-1 in your formula). Case 1 looks like this:
      $underset{c1}{}|underset{c2}{text{some balls}}|underset{c3}{text{some balls}}|underset{c4}{}$. Of course, under "some balls" we know we have 5 of them and
      c1 and c4 are chosen to have none. The problem then becomes: in how many ways can we write 5 has a sum of 2 nonnegative integers? Here, n=5 and k=1.



      Case 2 has c1=c4=1 so we have the question: in how many ways can we write 5-2=3 as a sum of 2 nonnegative integers (remaining balls)? Run through all the cases and add them up.



      Does this makes sense? Maybe there's a way easier way to solve this but hey, its an idea. What I'm saying is that my strategy would be to consider the actual proof and see what modifications need to be made






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        I haven't done combinatorics in a long time so I might not give you the correct answer but here are some thoughts:



        First off, perhaps you should clarify/consider if zeros are allowed. Let's suppose this is so. Think about how your formula can be justified combinatorially: using balls and walls. You can read that here: https://math.dartmouth.edu/archive/m68f09/public_html/m68l02.pdf



        I would think of a simpler case. writing 5 as a sum of 4 parts c1, c2, c3 and c4. My idea would then be that we would have 3 cases: c1=c4=0,1 or 2 (3 won't do it). Also notice that I'm using a multiple of 5 for a reason. I guess you would have to take the floor or ceiling function of 75/2.



        We're gonna have to use 3 walls (k-1 in your formula). Case 1 looks like this:
        $underset{c1}{}|underset{c2}{text{some balls}}|underset{c3}{text{some balls}}|underset{c4}{}$. Of course, under "some balls" we know we have 5 of them and
        c1 and c4 are chosen to have none. The problem then becomes: in how many ways can we write 5 has a sum of 2 nonnegative integers? Here, n=5 and k=1.



        Case 2 has c1=c4=1 so we have the question: in how many ways can we write 5-2=3 as a sum of 2 nonnegative integers (remaining balls)? Run through all the cases and add them up.



        Does this makes sense? Maybe there's a way easier way to solve this but hey, its an idea. What I'm saying is that my strategy would be to consider the actual proof and see what modifications need to be made






        share|cite|improve this answer











        $endgroup$



        I haven't done combinatorics in a long time so I might not give you the correct answer but here are some thoughts:



        First off, perhaps you should clarify/consider if zeros are allowed. Let's suppose this is so. Think about how your formula can be justified combinatorially: using balls and walls. You can read that here: https://math.dartmouth.edu/archive/m68f09/public_html/m68l02.pdf



        I would think of a simpler case. writing 5 as a sum of 4 parts c1, c2, c3 and c4. My idea would then be that we would have 3 cases: c1=c4=0,1 or 2 (3 won't do it). Also notice that I'm using a multiple of 5 for a reason. I guess you would have to take the floor or ceiling function of 75/2.



        We're gonna have to use 3 walls (k-1 in your formula). Case 1 looks like this:
        $underset{c1}{}|underset{c2}{text{some balls}}|underset{c3}{text{some balls}}|underset{c4}{}$. Of course, under "some balls" we know we have 5 of them and
        c1 and c4 are chosen to have none. The problem then becomes: in how many ways can we write 5 has a sum of 2 nonnegative integers? Here, n=5 and k=1.



        Case 2 has c1=c4=1 so we have the question: in how many ways can we write 5-2=3 as a sum of 2 nonnegative integers (remaining balls)? Run through all the cases and add them up.



        Does this makes sense? Maybe there's a way easier way to solve this but hey, its an idea. What I'm saying is that my strategy would be to consider the actual proof and see what modifications need to be made







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 20:55

























        answered Jan 22 at 20:49









        JayJay

        296




        296






























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