Dtjytyk

Is there a general formula for this?

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.

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

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
$$
{2 choose 2} + {3 choose 2} + {4 choose 2} + {5 choose 2} + {6 choose 2} +
{7 choose 2} + {8 choose 2}
$$