Number of positive unequal integer solution












0












$begingroup$



Number of positive unequal integer solution of



$x_{1}+x_{2}+x_{3}+x_{4}=20,$ is




Try: Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



Now for unequal integer solution, First we will find Total positive integer solution.



And subtract the cases where any three are equal and any two are equal and all are equal.



So Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



And number of positive integer solution in which all are equal i. e $x_{1}=x_{2}=x_{3}=x_{4}.$ Which is only one quadraplets.



Now if all any three are equal i. e $x_{1}=x_{2}=x_{3}$. Then $3x_{1}+x_{4}=20.$ we have $6times 4=24.$



If any two are equal. i. e $x_{1}=x_{2}.$



Then $2x_{1}+x_{2}+x_{3}=20.$ So we have $6times displaystyle sum^{9}_{k=1}binom{19-2k}{1}=6sum^{9}_{k=1}(19-k)=126times6=756.$



So Total number of positive nequal inteher solution $$binom{19}{3}-1-24-756=969-781=$$



But answer given as $552.$



Could some help me How to solve it, Thanks






combinatorics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    When counting the solutions where any 3 variables are equal, you included the solution with all 4 equal. In fact you included it 4 times. – I suppose you made a similar mistake for the solutions with 2 variables being equal.
    $endgroup$
    – Christoph
    Dec 7 '18 at 7:24










  • $begingroup$
    Are you looking for solutions where none of the $x_i$'s are equal or the ones where not all four are equal?
    $endgroup$
    – Anurag A
    Dec 7 '18 at 8:03
















0












$begingroup$



Number of positive unequal integer solution of



$x_{1}+x_{2}+x_{3}+x_{4}=20,$ is




Try: Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



Now for unequal integer solution, First we will find Total positive integer solution.



And subtract the cases where any three are equal and any two are equal and all are equal.



So Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



And number of positive integer solution in which all are equal i. e $x_{1}=x_{2}=x_{3}=x_{4}.$ Which is only one quadraplets.



Now if all any three are equal i. e $x_{1}=x_{2}=x_{3}$. Then $3x_{1}+x_{4}=20.$ we have $6times 4=24.$



If any two are equal. i. e $x_{1}=x_{2}.$



Then $2x_{1}+x_{2}+x_{3}=20.$ So we have $6times displaystyle sum^{9}_{k=1}binom{19-2k}{1}=6sum^{9}_{k=1}(19-k)=126times6=756.$



So Total number of positive nequal inteher solution $$binom{19}{3}-1-24-756=969-781=$$



But answer given as $552.$



Could some help me How to solve it, Thanks






combinatorics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    When counting the solutions where any 3 variables are equal, you included the solution with all 4 equal. In fact you included it 4 times. – I suppose you made a similar mistake for the solutions with 2 variables being equal.
    $endgroup$
    – Christoph
    Dec 7 '18 at 7:24










  • $begingroup$
    Are you looking for solutions where none of the $x_i$'s are equal or the ones where not all four are equal?
    $endgroup$
    – Anurag A
    Dec 7 '18 at 8:03














0












0








0





$begingroup$



Number of positive unequal integer solution of



$x_{1}+x_{2}+x_{3}+x_{4}=20,$ is




Try: Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



Now for unequal integer solution, First we will find Total positive integer solution.



And subtract the cases where any three are equal and any two are equal and all are equal.



So Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



And number of positive integer solution in which all are equal i. e $x_{1}=x_{2}=x_{3}=x_{4}.$ Which is only one quadraplets.



Now if all any three are equal i. e $x_{1}=x_{2}=x_{3}$. Then $3x_{1}+x_{4}=20.$ we have $6times 4=24.$



If any two are equal. i. e $x_{1}=x_{2}.$



Then $2x_{1}+x_{2}+x_{3}=20.$ So we have $6times displaystyle sum^{9}_{k=1}binom{19-2k}{1}=6sum^{9}_{k=1}(19-k)=126times6=756.$



So Total number of positive nequal inteher solution $$binom{19}{3}-1-24-756=969-781=$$



But answer given as $552.$



Could some help me How to solve it, Thanks






combinatorics







share|cite|improve this question











$endgroup$





Number of positive unequal integer solution of



$x_{1}+x_{2}+x_{3}+x_{4}=20,$ is




Try: Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



Now for unequal integer solution, First we will find Total positive integer solution.



And subtract the cases where any three are equal and any two are equal and all are equal.



So Total number of positive integer solution is



$displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method)



And number of positive integer solution in which all are equal i. e $x_{1}=x_{2}=x_{3}=x_{4}.$ Which is only one quadraplets.



Now if all any three are equal i. e $x_{1}=x_{2}=x_{3}$. Then $3x_{1}+x_{4}=20.$ we have $6times 4=24.$



If any two are equal. i. e $x_{1}=x_{2}.$



Then $2x_{1}+x_{2}+x_{3}=20.$ So we have $6times displaystyle sum^{9}_{k=1}binom{19-2k}{1}=6sum^{9}_{k=1}(19-k)=126times6=756.$



So Total number of positive nequal inteher solution $$binom{19}{3}-1-24-756=969-781=$$



But answer given as $552.$



Could some help me How to solve it, Thanks






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 7:46







DXT

















asked Dec 7 '18 at 7:11









DXTDXT

5,6662630




5,6662630








  • 2




    $begingroup$
    When counting the solutions where any 3 variables are equal, you included the solution with all 4 equal. In fact you included it 4 times. – I suppose you made a similar mistake for the solutions with 2 variables being equal.
    $endgroup$
    – Christoph
    Dec 7 '18 at 7:24










  • $begingroup$
    Are you looking for solutions where none of the $x_i$'s are equal or the ones where not all four are equal?
    $endgroup$
    – Anurag A
    Dec 7 '18 at 8:03














  • 2




    $begingroup$
    When counting the solutions where any 3 variables are equal, you included the solution with all 4 equal. In fact you included it 4 times. – I suppose you made a similar mistake for the solutions with 2 variables being equal.
    $endgroup$
    – Christoph
    Dec 7 '18 at 7:24










  • $begingroup$
    Are you looking for solutions where none of the $x_i$'s are equal or the ones where not all four are equal?
    $endgroup$
    – Anurag A
    Dec 7 '18 at 8:03








2




2




$begingroup$
When counting the solutions where any 3 variables are equal, you included the solution with all 4 equal. In fact you included it 4 times. – I suppose you made a similar mistake for the solutions with 2 variables being equal.
$endgroup$
– Christoph
Dec 7 '18 at 7:24




$begingroup$
When counting the solutions where any 3 variables are equal, you included the solution with all 4 equal. In fact you included it 4 times. – I suppose you made a similar mistake for the solutions with 2 variables being equal.
$endgroup$
– Christoph
Dec 7 '18 at 7:24












$begingroup$
Are you looking for solutions where none of the $x_i$'s are equal or the ones where not all four are equal?
$endgroup$
– Anurag A
Dec 7 '18 at 8:03




$begingroup$
Are you looking for solutions where none of the $x_i$'s are equal or the ones where not all four are equal?
$endgroup$
– Anurag A
Dec 7 '18 at 8:03










1 Answer
1






active

oldest

votes


















3












$begingroup$

Counting the positive integer solutions of $x_1+x_2+x_3+x_4=20$ with unequal parts, we may assume $x_1<x_2<x_3<x_4$ and multiply by $4!$ in the end.



Now letting $y_i = x_i - i$ for $i=1,2,3,4$ this translates to counting quadruplets $(y_1,y_2,y_3,y_4)$ of non-negative integers with $y_1le y_2le y_3le y_4$ and $y_1+y_2+y_3+y_4=10$. Hence, we are counting partitions of $10$ with at most $4$ parts. These are $23$:
begin{align*}
10 &= 10 \
&= 9+1 = 8+2 = 7+3 = 6+4 = 5+5 \
&= 8+1+1 = 7+2+1 = 6+3+1 = 6+2+2 \
&= 5+4+1 = 5+3+2 = 4+4+2 = 4+3+3 \
&= 7+1+1+1 \&= 6+2+1+1 \&= 5+3+1+1 = 5+2+2+1 \
&= 4+4+1+1 = 4+3+2+1 = 4+2+2+2 \
&= 3+3+3+1 = 3+3+2+2.
end{align*}



Hence, there are $23cdot 4!=552$ solutions to your problem.




You could also go one more step and let $z_1=y_1$, $z_i=y_i-y_{i-1}$ for $i=2,3,4$ to translate this to counting the non-negative integer solutions of $4z_1+3z_2+2z_3+z_4=10$. (This amounts to conjugating the partitions and count those with parts of size at most $4$)






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    $begingroup$

    Counting the positive integer solutions of $x_1+x_2+x_3+x_4=20$ with unequal parts, we may assume $x_1<x_2<x_3<x_4$ and multiply by $4!$ in the end.



    Now letting $y_i = x_i - i$ for $i=1,2,3,4$ this translates to counting quadruplets $(y_1,y_2,y_3,y_4)$ of non-negative integers with $y_1le y_2le y_3le y_4$ and $y_1+y_2+y_3+y_4=10$. Hence, we are counting partitions of $10$ with at most $4$ parts. These are $23$:
    begin{align*}
    10 &= 10 \
    &= 9+1 = 8+2 = 7+3 = 6+4 = 5+5 \
    &= 8+1+1 = 7+2+1 = 6+3+1 = 6+2+2 \
    &= 5+4+1 = 5+3+2 = 4+4+2 = 4+3+3 \
    &= 7+1+1+1 \&= 6+2+1+1 \&= 5+3+1+1 = 5+2+2+1 \
    &= 4+4+1+1 = 4+3+2+1 = 4+2+2+2 \
    &= 3+3+3+1 = 3+3+2+2.
    end{align*}



    Hence, there are $23cdot 4!=552$ solutions to your problem.




    You could also go one more step and let $z_1=y_1$, $z_i=y_i-y_{i-1}$ for $i=2,3,4$ to translate this to counting the non-negative integer solutions of $4z_1+3z_2+2z_3+z_4=10$. (This amounts to conjugating the partitions and count those with parts of size at most $4$)






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Counting the positive integer solutions of $x_1+x_2+x_3+x_4=20$ with unequal parts, we may assume $x_1<x_2<x_3<x_4$ and multiply by $4!$ in the end.



      Now letting $y_i = x_i - i$ for $i=1,2,3,4$ this translates to counting quadruplets $(y_1,y_2,y_3,y_4)$ of non-negative integers with $y_1le y_2le y_3le y_4$ and $y_1+y_2+y_3+y_4=10$. Hence, we are counting partitions of $10$ with at most $4$ parts. These are $23$:
      begin{align*}
      10 &= 10 \
      &= 9+1 = 8+2 = 7+3 = 6+4 = 5+5 \
      &= 8+1+1 = 7+2+1 = 6+3+1 = 6+2+2 \
      &= 5+4+1 = 5+3+2 = 4+4+2 = 4+3+3 \
      &= 7+1+1+1 \&= 6+2+1+1 \&= 5+3+1+1 = 5+2+2+1 \
      &= 4+4+1+1 = 4+3+2+1 = 4+2+2+2 \
      &= 3+3+3+1 = 3+3+2+2.
      end{align*}



      Hence, there are $23cdot 4!=552$ solutions to your problem.




      You could also go one more step and let $z_1=y_1$, $z_i=y_i-y_{i-1}$ for $i=2,3,4$ to translate this to counting the non-negative integer solutions of $4z_1+3z_2+2z_3+z_4=10$. (This amounts to conjugating the partitions and count those with parts of size at most $4$)






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Counting the positive integer solutions of $x_1+x_2+x_3+x_4=20$ with unequal parts, we may assume $x_1<x_2<x_3<x_4$ and multiply by $4!$ in the end.



        Now letting $y_i = x_i - i$ for $i=1,2,3,4$ this translates to counting quadruplets $(y_1,y_2,y_3,y_4)$ of non-negative integers with $y_1le y_2le y_3le y_4$ and $y_1+y_2+y_3+y_4=10$. Hence, we are counting partitions of $10$ with at most $4$ parts. These are $23$:
        begin{align*}
        10 &= 10 \
        &= 9+1 = 8+2 = 7+3 = 6+4 = 5+5 \
        &= 8+1+1 = 7+2+1 = 6+3+1 = 6+2+2 \
        &= 5+4+1 = 5+3+2 = 4+4+2 = 4+3+3 \
        &= 7+1+1+1 \&= 6+2+1+1 \&= 5+3+1+1 = 5+2+2+1 \
        &= 4+4+1+1 = 4+3+2+1 = 4+2+2+2 \
        &= 3+3+3+1 = 3+3+2+2.
        end{align*}



        Hence, there are $23cdot 4!=552$ solutions to your problem.




        You could also go one more step and let $z_1=y_1$, $z_i=y_i-y_{i-1}$ for $i=2,3,4$ to translate this to counting the non-negative integer solutions of $4z_1+3z_2+2z_3+z_4=10$. (This amounts to conjugating the partitions and count those with parts of size at most $4$)






        share|cite|improve this answer











        $endgroup$



        Counting the positive integer solutions of $x_1+x_2+x_3+x_4=20$ with unequal parts, we may assume $x_1<x_2<x_3<x_4$ and multiply by $4!$ in the end.



        Now letting $y_i = x_i - i$ for $i=1,2,3,4$ this translates to counting quadruplets $(y_1,y_2,y_3,y_4)$ of non-negative integers with $y_1le y_2le y_3le y_4$ and $y_1+y_2+y_3+y_4=10$. Hence, we are counting partitions of $10$ with at most $4$ parts. These are $23$:
        begin{align*}
        10 &= 10 \
        &= 9+1 = 8+2 = 7+3 = 6+4 = 5+5 \
        &= 8+1+1 = 7+2+1 = 6+3+1 = 6+2+2 \
        &= 5+4+1 = 5+3+2 = 4+4+2 = 4+3+3 \
        &= 7+1+1+1 \&= 6+2+1+1 \&= 5+3+1+1 = 5+2+2+1 \
        &= 4+4+1+1 = 4+3+2+1 = 4+2+2+2 \
        &= 3+3+3+1 = 3+3+2+2.
        end{align*}



        Hence, there are $23cdot 4!=552$ solutions to your problem.




        You could also go one more step and let $z_1=y_1$, $z_i=y_i-y_{i-1}$ for $i=2,3,4$ to translate this to counting the non-negative integer solutions of $4z_1+3z_2+2z_3+z_4=10$. (This amounts to conjugating the partitions and count those with parts of size at most $4$)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 8:16

























        answered Dec 7 '18 at 8:07









        ChristophChristoph

        11.9k1642




        11.9k1642






























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