Number of nonnegative integer solutions to $x_1+x_2+x_3le10$ with $x_1 ge 1 , x_2ge3$Solve...

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Number of nonnegative integer solutions to $x_1+x_2+x_3le10$ with $x_1 ge 1 , x_2ge3$


Solve $int_{0}^{1}frac{1}{1+x^6} dx$Number of integer solutions to the equation $x_1 + x_2 + x_3 = 28$ with rangesDetermine the Number of Integer Solutions $x_1 + x_2 + x_3 + x_4 = 32$ with restrictionsProving that $max(x_1, x_2, x_3) = x_1 + x_2 + x_3 - min(x_1, x_2) - min(x_1, x_3) - min(x_2, x_3) + min(x_1, x_2, x_3)$Generalized way to solve $x_1 + x_2 + x_3 = c$ with the constraint $x_1 > x_2 > x_3$?positive integer solutions to $x_1 cdot x_2 cdot x_3=500$Number of non-negative integers solutions of $x_1 + x_2 + x_3 + x_4 + x_5 = 10$ when $x_1 = x_2$ and when $x_1 > x_2$How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with restrictions on $x_1,x_2,x_3,x_4$?Number of solutions $(x_1)(x_2)(x_3)(x_4) = 2016$How many nonnegative integer solutions are there to the equation $x_1+x_2+x_3+x_4 = 18$ with restrictionsAll possible solutions for $x_1+x_2+x_3=25$ with conditions for $x_1, x_2$ and $x_3$ ranges













0












$begingroup$


Is there a general formula for this?






combinatorics discrete-mathematics elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, a small modification of Stars and Bars works. Do your natural numbers include $0$? Probably, else the problem would not say $x_1ge 1$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 22:59












  • $begingroup$
    Yes $N$ includes $0$. I am guessing the answer is ${12-1-3 choose 2 }$ but i have trouble with the intuition.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:08
















0












$begingroup$


Is there a general formula for this?






combinatorics discrete-mathematics elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, a small modification of Stars and Bars works. Do your natural numbers include $0$? Probably, else the problem would not say $x_1ge 1$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 22:59












  • $begingroup$
    Yes $N$ includes $0$. I am guessing the answer is ${12-1-3 choose 2 }$ but i have trouble with the intuition.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:08














0












0








0


1



$begingroup$


Is there a general formula for this?






combinatorics discrete-mathematics elementary-set-theory







share|cite|improve this question











$endgroup$




Is there a general formula for this?






combinatorics discrete-mathematics elementary-set-theory




combinatorics discrete-mathematics elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 31 '16 at 23:20









Rodrigo de Azevedo

13k41960




13k41960










asked Jul 31 '16 at 22:57









SoloNasusSoloNasus

420216




420216












  • $begingroup$
    Yes, a small modification of Stars and Bars works. Do your natural numbers include $0$? Probably, else the problem would not say $x_1ge 1$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 22:59












  • $begingroup$
    Yes $N$ includes $0$. I am guessing the answer is ${12-1-3 choose 2 }$ but i have trouble with the intuition.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:08


















  • $begingroup$
    Yes, a small modification of Stars and Bars works. Do your natural numbers include $0$? Probably, else the problem would not say $x_1ge 1$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 22:59












  • $begingroup$
    Yes $N$ includes $0$. I am guessing the answer is ${12-1-3 choose 2 }$ but i have trouble with the intuition.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:08
















$begingroup$
Yes, a small modification of Stars and Bars works. Do your natural numbers include $0$? Probably, else the problem would not say $x_1ge 1$.
$endgroup$
– André Nicolas
Jul 31 '16 at 22:59






$begingroup$
Yes, a small modification of Stars and Bars works. Do your natural numbers include $0$? Probably, else the problem would not say $x_1ge 1$.
$endgroup$
– André Nicolas
Jul 31 '16 at 22:59














$begingroup$
Yes $N$ includes $0$. I am guessing the answer is ${12-1-3 choose 2 }$ but i have trouble with the intuition.
$endgroup$
– SoloNasus
Jul 31 '16 at 23:08




$begingroup$
Yes $N$ includes $0$. I am guessing the answer is ${12-1-3 choose 2 }$ but i have trouble with the intuition.
$endgroup$
– SoloNasus
Jul 31 '16 at 23:08










2 Answers
2






active

oldest

votes


















4












$begingroup$

I am giving away up to $10$ identical candies among $3$ kids, with the provisos that Kid 1 gets at least $1,$ and Kid 2 gets at least $3$.



Give $1$ candy to Kid 1, and $3$ to Kid 2. Now we are distributing up to $6$ candies among $3$ kids. Invent a fourth kid, Kid 4, who will get all the "leftover" candies.



So the number of ways to carry out our task is the number of ways to distribute all $6$ candies among these $4$ kids. Standard Stars and Bars shows that there are $binom{6+4-1}{4-1}$ ways to do this.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:25












  • $begingroup$
    @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:32












  • $begingroup$
    @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:41










  • $begingroup$
    Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:47












  • $begingroup$
    @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:54



















0












$begingroup$

$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}_{#1}}
newcommand{mc}[1]{,mathcal{#1}}
newcommand{mrm}[1]{,mathrm{#1}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$






  1. $ds{sum_{x_{1} = color{#f00}{1}}^{infty}
    sum_{x_{2} = color{#f00}{3}}^{infty}
    sum_{x_{3} = color{#f00}{0}}^{infty}
    delta_{x_{1} + x_{2} + x_{3},,, s},,,}$     counts the number of triplets
    $ds{pars{x_{1},x_{2},x_{3}}}$ with $ds{x_{1} + x_{2} + x_{3} = s}$ because the Kronecker Delta
    $ds{delta_{x_{1} + x_{2} + x_{3},,, s},,,}$ is equal to $ds{1}$ when the above condition is satisfied and $ds{0}$ otherwise.

  2. The next sum
    $ds{sum_{s = 4}^{10}cdots}$ completes the task by collecting the total sums
    $ds{s = 4,5,6,7,8,9,10}$.

  3. The evaluation is straightforward with a well known Kronecker Delta integral representation which allows to sum over independent
    $ds{braces{x_{i}, i = 1,2,3}}$.

  4. Sums over $ds{braces{x_{i}, i,1,2,3}}$ can be extended to $ds{infty}$ because the Kronecker Delta doesn't allow the case
    $ds{x_{1} + x_{2} + x_{3} > s}$
    $ds{pars{~it vanishes out in such cases~}}$.




begin{align}
&sum_{s = color{#f00}{4}}^{color{#f00}{10}}sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
sum_{x_{3} = color{#f00}{0}}^{infty}delta_{x_{1} + x_{2} + x_{3},,, s}
,,,=,,,
sum_{s = color{#f00}{4}}^{color{#f00}{10}}
sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
sum_{x_{3} = color{#f00}{0}}^{infty}
,,oint_{verts{z} = color{#f00}{1^{-}}},,,
{1 over z^{s + 1 - x_{1} - x_{2} - x_{3}}},,,{dd z over 2piic}
\[5mm] = &
sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
pars{sum_{x_{1} = color{#f00}{1}}^{infty}z^{x_{1}}}
pars{sum_{x_{2} = color{#f00}{3}}^{infty}z^{x_{2}}}
pars{sum_{x_{3} = color{#f00}{0}}^{infty}z^{x_{3}}}
,{dd z over 2piic}
\[5mm] & =
sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
{z over 1 - z}{z^{3} over 1 - z}{1 over 1 - z},{dd z over 2piic} =
sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
oint_{verts{z} = color{#f00}{1^{-}}},,
{1 over z^{s - 3},,pars{1 - z}^{3}},{dd z over 2piic}
\[5mm] & =
sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
sum_{k = 0}^{infty}{-3 choose k}pars{-1}^{k}
oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s - 3 - k}}
,,{dd z over 2piic}
\[5mm] & =
sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
sum_{k = 0}^{infty}{3 + k - 1choose k}pars{-1}^{k}pars{-1}^{k},
delta_{s - k - 3,1}
\[5mm] & =
sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
sum_{k = 0}^{infty}half,pars{k + 2}pars{k + 1}
delta_{k,s - 4}
\[5mm] & =
halfsum_{s = 0}^{6}pars{s + 2}pars{s + 1} = color{#f00}{84}quad
pars{~Longleftarrow ul{final result}~}
end{align}




which is equivalent to
$$
{2 choose 2} + {3 choose 2} + {4 choose 2} + {5 choose 2} + {6 choose 2} +
{7 choose 2} + {8 choose 2}
$$







share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    I am giving away up to $10$ identical candies among $3$ kids, with the provisos that Kid 1 gets at least $1,$ and Kid 2 gets at least $3$.



    Give $1$ candy to Kid 1, and $3$ to Kid 2. Now we are distributing up to $6$ candies among $3$ kids. Invent a fourth kid, Kid 4, who will get all the "leftover" candies.



    So the number of ways to carry out our task is the number of ways to distribute all $6$ candies among these $4$ kids. Standard Stars and Bars shows that there are $binom{6+4-1}{4-1}$ ways to do this.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:25












    • $begingroup$
      @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:32












    • $begingroup$
      @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:41










    • $begingroup$
      Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:47












    • $begingroup$
      @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:54
















    4












    $begingroup$

    I am giving away up to $10$ identical candies among $3$ kids, with the provisos that Kid 1 gets at least $1,$ and Kid 2 gets at least $3$.



    Give $1$ candy to Kid 1, and $3$ to Kid 2. Now we are distributing up to $6$ candies among $3$ kids. Invent a fourth kid, Kid 4, who will get all the "leftover" candies.



    So the number of ways to carry out our task is the number of ways to distribute all $6$ candies among these $4$ kids. Standard Stars and Bars shows that there are $binom{6+4-1}{4-1}$ ways to do this.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:25












    • $begingroup$
      @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:32












    • $begingroup$
      @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:41










    • $begingroup$
      Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:47












    • $begingroup$
      @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:54














    4












    4








    4





    $begingroup$

    I am giving away up to $10$ identical candies among $3$ kids, with the provisos that Kid 1 gets at least $1,$ and Kid 2 gets at least $3$.



    Give $1$ candy to Kid 1, and $3$ to Kid 2. Now we are distributing up to $6$ candies among $3$ kids. Invent a fourth kid, Kid 4, who will get all the "leftover" candies.



    So the number of ways to carry out our task is the number of ways to distribute all $6$ candies among these $4$ kids. Standard Stars and Bars shows that there are $binom{6+4-1}{4-1}$ ways to do this.






    share|cite|improve this answer









    $endgroup$



    I am giving away up to $10$ identical candies among $3$ kids, with the provisos that Kid 1 gets at least $1,$ and Kid 2 gets at least $3$.



    Give $1$ candy to Kid 1, and $3$ to Kid 2. Now we are distributing up to $6$ candies among $3$ kids. Invent a fourth kid, Kid 4, who will get all the "leftover" candies.



    So the number of ways to carry out our task is the number of ways to distribute all $6$ candies among these $4$ kids. Standard Stars and Bars shows that there are $binom{6+4-1}{4-1}$ ways to do this.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 31 '16 at 23:17









    André NicolasAndré Nicolas

    454k36430817




    454k36430817












    • $begingroup$
      Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:25












    • $begingroup$
      @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:32












    • $begingroup$
      @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:41










    • $begingroup$
      Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:47












    • $begingroup$
      @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:54


















    • $begingroup$
      Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:25












    • $begingroup$
      @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:32












    • $begingroup$
      @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:41










    • $begingroup$
      Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
      $endgroup$
      – SoloNasus
      Jul 31 '16 at 23:47












    • $begingroup$
      @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
      $endgroup$
      – André Nicolas
      Jul 31 '16 at 23:54
















    $begingroup$
    Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:25






    $begingroup$
    Oh, it seems like i have made a typo in the title. What I meant to write was $x_1+x_2+x_3=10$. Regardless your solution helped me understand the thinking behind it.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:25














    $begingroup$
    @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:32






    $begingroup$
    @ТеодорДяков: Might as well keep it that way, now you know how to deal with the twist $le n$. For $=10$, Stars and Bars gives $binom{6+3-1}{3-1}$. I did not deal with the Stars and Bars intuition in my answer, since it seemed the main task was to explain how to deal with $le$.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:32














    $begingroup$
    @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:41




    $begingroup$
    @ТеодорДяков: I am not quite happy with the "standard" Stars and Bars intuition. To me, distributing $n$ candies among $k$ kids so that each gets at least $1$ is clearly $binom{n-1}{k-1}$ (we choose $k-1$ places from $n-1$ to place separators). Then if $0$ is allowed, I think of it as distributing $n+k$ candies among $k$ kids, with each getting at least $1$ (which can be done in $binom{n+k-1}{k-1}$ ways) and then taking a candy away from each kid.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:41












    $begingroup$
    Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:47






    $begingroup$
    Yeah i knew about the Stars and Bars, the problem was $x_1ge1, x_2ge 3$ but it seems like it can be reduced to $z+y+x_3ge6 , z=x_1-1 , y=x_2-3$ and then use standard Stars and Bars.
    $endgroup$
    – SoloNasus
    Jul 31 '16 at 23:47














    $begingroup$
    @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:54




    $begingroup$
    @ТеодорДяков: The $ge 1$, $ge 3$ are no problem. You can either make a change of variable, as in your comment above, or explain it, as in the post, in terms of giving preliminary gifts of $1$ and $3$ to Kids 1 and 2. Depends on whether algebra or concrete view of the situation is more natural to you.
    $endgroup$
    – André Nicolas
    Jul 31 '16 at 23:54











    0












    $begingroup$

    $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{half}{{1 over 2}}
    newcommand{ic}{mathrm{i}}
    newcommand{iff}{Longleftrightarrow}
    newcommand{imp}{Longrightarrow}
    newcommand{Li}[1]{,mathrm{Li}_{#1}}
    newcommand{mc}[1]{,mathcal{#1}}
    newcommand{mrm}[1]{,mathrm{#1}}
    newcommand{ol}[1]{overline{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{ul}[1]{underline{#1}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$






    1. $ds{sum_{x_{1} = color{#f00}{1}}^{infty}
      sum_{x_{2} = color{#f00}{3}}^{infty}
      sum_{x_{3} = color{#f00}{0}}^{infty}
      delta_{x_{1} + x_{2} + x_{3},,, s},,,}$       counts the number of triplets
      $ds{pars{x_{1},x_{2},x_{3}}}$ with $ds{x_{1} + x_{2} + x_{3} = s}$ because the Kronecker Delta
      $ds{delta_{x_{1} + x_{2} + x_{3},,, s},,,}$ is equal to $ds{1}$ when the above condition is satisfied and $ds{0}$ otherwise.

    2. The next sum
      $ds{sum_{s = 4}^{10}cdots}$ completes the task by collecting the total sums
      $ds{s = 4,5,6,7,8,9,10}$.

    3. The evaluation is straightforward with a well known Kronecker Delta integral representation which allows to sum over independent
      $ds{braces{x_{i}, i = 1,2,3}}$.

    4. Sums over $ds{braces{x_{i}, i,1,2,3}}$ can be extended to $ds{infty}$ because the Kronecker Delta doesn't allow the case
      $ds{x_{1} + x_{2} + x_{3} > s}$
      $ds{pars{~it vanishes out in such cases~}}$.




    begin{align}
    &sum_{s = color{#f00}{4}}^{color{#f00}{10}}sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
    sum_{x_{3} = color{#f00}{0}}^{infty}delta_{x_{1} + x_{2} + x_{3},,, s}
    ,,,=,,,
    sum_{s = color{#f00}{4}}^{color{#f00}{10}}
    sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
    sum_{x_{3} = color{#f00}{0}}^{infty}
    ,,oint_{verts{z} = color{#f00}{1^{-}}},,,
    {1 over z^{s + 1 - x_{1} - x_{2} - x_{3}}},,,{dd z over 2piic}
    \[5mm] = &
    sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
    oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
    pars{sum_{x_{1} = color{#f00}{1}}^{infty}z^{x_{1}}}
    pars{sum_{x_{2} = color{#f00}{3}}^{infty}z^{x_{2}}}
    pars{sum_{x_{3} = color{#f00}{0}}^{infty}z^{x_{3}}}
    ,{dd z over 2piic}
    \[5mm] & =
    sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
    oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
    {z over 1 - z}{z^{3} over 1 - z}{1 over 1 - z},{dd z over 2piic} =
    sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
    oint_{verts{z} = color{#f00}{1^{-}}},,
    {1 over z^{s - 3},,pars{1 - z}^{3}},{dd z over 2piic}
    \[5mm] & =
    sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
    sum_{k = 0}^{infty}{-3 choose k}pars{-1}^{k}
    oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s - 3 - k}}
    ,,{dd z over 2piic}
    \[5mm] & =
    sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
    sum_{k = 0}^{infty}{3 + k - 1choose k}pars{-1}^{k}pars{-1}^{k},
    delta_{s - k - 3,1}
    \[5mm] & =
    sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
    sum_{k = 0}^{infty}half,pars{k + 2}pars{k + 1}
    delta_{k,s - 4}
    \[5mm] & =
    halfsum_{s = 0}^{6}pars{s + 2}pars{s + 1} = color{#f00}{84}quad
    pars{~Longleftarrow ul{final result}~}
    end{align}




    which is equivalent to
    $$
    {2 choose 2} + {3 choose 2} + {4 choose 2} + {5 choose 2} + {6 choose 2} +
    {7 choose 2} + {8 choose 2}
    $$







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{half}{{1 over 2}}
      newcommand{ic}{mathrm{i}}
      newcommand{iff}{Longleftrightarrow}
      newcommand{imp}{Longrightarrow}
      newcommand{Li}[1]{,mathrm{Li}_{#1}}
      newcommand{mc}[1]{,mathcal{#1}}
      newcommand{mrm}[1]{,mathrm{#1}}
      newcommand{ol}[1]{overline{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{ul}[1]{underline{#1}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$






      1. $ds{sum_{x_{1} = color{#f00}{1}}^{infty}
        sum_{x_{2} = color{#f00}{3}}^{infty}
        sum_{x_{3} = color{#f00}{0}}^{infty}
        delta_{x_{1} + x_{2} + x_{3},,, s},,,}$         counts the number of triplets
        $ds{pars{x_{1},x_{2},x_{3}}}$ with $ds{x_{1} + x_{2} + x_{3} = s}$ because the Kronecker Delta
        $ds{delta_{x_{1} + x_{2} + x_{3},,, s},,,}$ is equal to $ds{1}$ when the above condition is satisfied and $ds{0}$ otherwise.

      2. The next sum
        $ds{sum_{s = 4}^{10}cdots}$ completes the task by collecting the total sums
        $ds{s = 4,5,6,7,8,9,10}$.

      3. The evaluation is straightforward with a well known Kronecker Delta integral representation which allows to sum over independent
        $ds{braces{x_{i}, i = 1,2,3}}$.

      4. Sums over $ds{braces{x_{i}, i,1,2,3}}$ can be extended to $ds{infty}$ because the Kronecker Delta doesn't allow the case
        $ds{x_{1} + x_{2} + x_{3} > s}$
        $ds{pars{~it vanishes out in such cases~}}$.




      begin{align}
      &sum_{s = color{#f00}{4}}^{color{#f00}{10}}sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
      sum_{x_{3} = color{#f00}{0}}^{infty}delta_{x_{1} + x_{2} + x_{3},,, s}
      ,,,=,,,
      sum_{s = color{#f00}{4}}^{color{#f00}{10}}
      sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
      sum_{x_{3} = color{#f00}{0}}^{infty}
      ,,oint_{verts{z} = color{#f00}{1^{-}}},,,
      {1 over z^{s + 1 - x_{1} - x_{2} - x_{3}}},,,{dd z over 2piic}
      \[5mm] = &
      sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
      oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
      pars{sum_{x_{1} = color{#f00}{1}}^{infty}z^{x_{1}}}
      pars{sum_{x_{2} = color{#f00}{3}}^{infty}z^{x_{2}}}
      pars{sum_{x_{3} = color{#f00}{0}}^{infty}z^{x_{3}}}
      ,{dd z over 2piic}
      \[5mm] & =
      sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
      oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
      {z over 1 - z}{z^{3} over 1 - z}{1 over 1 - z},{dd z over 2piic} =
      sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
      oint_{verts{z} = color{#f00}{1^{-}}},,
      {1 over z^{s - 3},,pars{1 - z}^{3}},{dd z over 2piic}
      \[5mm] & =
      sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
      sum_{k = 0}^{infty}{-3 choose k}pars{-1}^{k}
      oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s - 3 - k}}
      ,,{dd z over 2piic}
      \[5mm] & =
      sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
      sum_{k = 0}^{infty}{3 + k - 1choose k}pars{-1}^{k}pars{-1}^{k},
      delta_{s - k - 3,1}
      \[5mm] & =
      sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
      sum_{k = 0}^{infty}half,pars{k + 2}pars{k + 1}
      delta_{k,s - 4}
      \[5mm] & =
      halfsum_{s = 0}^{6}pars{s + 2}pars{s + 1} = color{#f00}{84}quad
      pars{~Longleftarrow ul{final result}~}
      end{align}




      which is equivalent to
      $$
      {2 choose 2} + {3 choose 2} + {4 choose 2} + {5 choose 2} + {6 choose 2} +
      {7 choose 2} + {8 choose 2}
      $$







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{half}{{1 over 2}}
        newcommand{ic}{mathrm{i}}
        newcommand{iff}{Longleftrightarrow}
        newcommand{imp}{Longrightarrow}
        newcommand{Li}[1]{,mathrm{Li}_{#1}}
        newcommand{mc}[1]{,mathcal{#1}}
        newcommand{mrm}[1]{,mathrm{#1}}
        newcommand{ol}[1]{overline{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{ul}[1]{underline{#1}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$






        1. $ds{sum_{x_{1} = color{#f00}{1}}^{infty}
          sum_{x_{2} = color{#f00}{3}}^{infty}
          sum_{x_{3} = color{#f00}{0}}^{infty}
          delta_{x_{1} + x_{2} + x_{3},,, s},,,}$           counts the number of triplets
          $ds{pars{x_{1},x_{2},x_{3}}}$ with $ds{x_{1} + x_{2} + x_{3} = s}$ because the Kronecker Delta
          $ds{delta_{x_{1} + x_{2} + x_{3},,, s},,,}$ is equal to $ds{1}$ when the above condition is satisfied and $ds{0}$ otherwise.

        2. The next sum
          $ds{sum_{s = 4}^{10}cdots}$ completes the task by collecting the total sums
          $ds{s = 4,5,6,7,8,9,10}$.

        3. The evaluation is straightforward with a well known Kronecker Delta integral representation which allows to sum over independent
          $ds{braces{x_{i}, i = 1,2,3}}$.

        4. Sums over $ds{braces{x_{i}, i,1,2,3}}$ can be extended to $ds{infty}$ because the Kronecker Delta doesn't allow the case
          $ds{x_{1} + x_{2} + x_{3} > s}$
          $ds{pars{~it vanishes out in such cases~}}$.




        begin{align}
        &sum_{s = color{#f00}{4}}^{color{#f00}{10}}sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
        sum_{x_{3} = color{#f00}{0}}^{infty}delta_{x_{1} + x_{2} + x_{3},,, s}
        ,,,=,,,
        sum_{s = color{#f00}{4}}^{color{#f00}{10}}
        sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
        sum_{x_{3} = color{#f00}{0}}^{infty}
        ,,oint_{verts{z} = color{#f00}{1^{-}}},,,
        {1 over z^{s + 1 - x_{1} - x_{2} - x_{3}}},,,{dd z over 2piic}
        \[5mm] = &
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
        pars{sum_{x_{1} = color{#f00}{1}}^{infty}z^{x_{1}}}
        pars{sum_{x_{2} = color{#f00}{3}}^{infty}z^{x_{2}}}
        pars{sum_{x_{3} = color{#f00}{0}}^{infty}z^{x_{3}}}
        ,{dd z over 2piic}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
        {z over 1 - z}{z^{3} over 1 - z}{1 over 1 - z},{dd z over 2piic} =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        oint_{verts{z} = color{#f00}{1^{-}}},,
        {1 over z^{s - 3},,pars{1 - z}^{3}},{dd z over 2piic}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        sum_{k = 0}^{infty}{-3 choose k}pars{-1}^{k}
        oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s - 3 - k}}
        ,,{dd z over 2piic}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        sum_{k = 0}^{infty}{3 + k - 1choose k}pars{-1}^{k}pars{-1}^{k},
        delta_{s - k - 3,1}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        sum_{k = 0}^{infty}half,pars{k + 2}pars{k + 1}
        delta_{k,s - 4}
        \[5mm] & =
        halfsum_{s = 0}^{6}pars{s + 2}pars{s + 1} = color{#f00}{84}quad
        pars{~Longleftarrow ul{final result}~}
        end{align}




        which is equivalent to
        $$
        {2 choose 2} + {3 choose 2} + {4 choose 2} + {5 choose 2} + {6 choose 2} +
        {7 choose 2} + {8 choose 2}
        $$







        share|cite|improve this answer











        $endgroup$



        $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{half}{{1 over 2}}
        newcommand{ic}{mathrm{i}}
        newcommand{iff}{Longleftrightarrow}
        newcommand{imp}{Longrightarrow}
        newcommand{Li}[1]{,mathrm{Li}_{#1}}
        newcommand{mc}[1]{,mathcal{#1}}
        newcommand{mrm}[1]{,mathrm{#1}}
        newcommand{ol}[1]{overline{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{ul}[1]{underline{#1}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$






        1. $ds{sum_{x_{1} = color{#f00}{1}}^{infty}
          sum_{x_{2} = color{#f00}{3}}^{infty}
          sum_{x_{3} = color{#f00}{0}}^{infty}
          delta_{x_{1} + x_{2} + x_{3},,, s},,,}$           counts the number of triplets
          $ds{pars{x_{1},x_{2},x_{3}}}$ with $ds{x_{1} + x_{2} + x_{3} = s}$ because the Kronecker Delta
          $ds{delta_{x_{1} + x_{2} + x_{3},,, s},,,}$ is equal to $ds{1}$ when the above condition is satisfied and $ds{0}$ otherwise.

        2. The next sum
          $ds{sum_{s = 4}^{10}cdots}$ completes the task by collecting the total sums
          $ds{s = 4,5,6,7,8,9,10}$.

        3. The evaluation is straightforward with a well known Kronecker Delta integral representation which allows to sum over independent
          $ds{braces{x_{i}, i = 1,2,3}}$.

        4. Sums over $ds{braces{x_{i}, i,1,2,3}}$ can be extended to $ds{infty}$ because the Kronecker Delta doesn't allow the case
          $ds{x_{1} + x_{2} + x_{3} > s}$
          $ds{pars{~it vanishes out in such cases~}}$.




        begin{align}
        &sum_{s = color{#f00}{4}}^{color{#f00}{10}}sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
        sum_{x_{3} = color{#f00}{0}}^{infty}delta_{x_{1} + x_{2} + x_{3},,, s}
        ,,,=,,,
        sum_{s = color{#f00}{4}}^{color{#f00}{10}}
        sum_{x_{1} = color{#f00}{1}}^{infty}sum_{x_{2} = color{#f00}{3}}^{infty}
        sum_{x_{3} = color{#f00}{0}}^{infty}
        ,,oint_{verts{z} = color{#f00}{1^{-}}},,,
        {1 over z^{s + 1 - x_{1} - x_{2} - x_{3}}},,,{dd z over 2piic}
        \[5mm] = &
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
        pars{sum_{x_{1} = color{#f00}{1}}^{infty}z^{x_{1}}}
        pars{sum_{x_{2} = color{#f00}{3}}^{infty}z^{x_{2}}}
        pars{sum_{x_{3} = color{#f00}{0}}^{infty}z^{x_{3}}}
        ,{dd z over 2piic}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s + 1}}
        {z over 1 - z}{z^{3} over 1 - z}{1 over 1 - z},{dd z over 2piic} =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        oint_{verts{z} = color{#f00}{1^{-}}},,
        {1 over z^{s - 3},,pars{1 - z}^{3}},{dd z over 2piic}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        sum_{k = 0}^{infty}{-3 choose k}pars{-1}^{k}
        oint_{verts{z} = color{#f00}{1^{-}}},,{1 over z^{s - 3 - k}}
        ,,{dd z over 2piic}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        sum_{k = 0}^{infty}{3 + k - 1choose k}pars{-1}^{k}pars{-1}^{k},
        delta_{s - k - 3,1}
        \[5mm] & =
        sum_{s = color{#f00}{4}}^{color{#f00}{10}},,
        sum_{k = 0}^{infty}half,pars{k + 2}pars{k + 1}
        delta_{k,s - 4}
        \[5mm] & =
        halfsum_{s = 0}^{6}pars{s + 2}pars{s + 1} = color{#f00}{84}quad
        pars{~Longleftarrow ul{final result}~}
        end{align}




        which is equivalent to
        $$
        {2 choose 2} + {3 choose 2} + {4 choose 2} + {5 choose 2} + {6 choose 2} +
        {7 choose 2} + {8 choose 2}
        $$








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        edited Mar 9 at 2:37

























        answered Aug 3 '16 at 2:26









        Felix MarinFelix Marin

        68.4k7109144




        68.4k7109144






























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