Mathematics Combinatorics If I have 10 BALLS and 3 boxes, how many possible number of solutions are there with a maximum number of 9 and minimum number of 1? numbers cannot be repeated.

Mathematically, we are finding the number of solutions of $x_1+x_2+x_3=10$ where $1\leq x_1,x_2,x_3\leq9$ and $x_1,x_2,x_3$ are integers.
Straight enumeration is the way to go here, but you can simplify a little I think.    Since you don't allow repeats, this is the same as $6$ times the number of ways to write $x_1 + x_2 + x_3 = 10$ with $x_i$ strictly increasing and greater than or equal to $1$.  Subtracting $i$ from $x_i$, we see that's the same as counting the number of $x_1 + x_2 + x_3 = 4$ with the $x_i$ weakly increasing and greater than $0$.  It's pretty quick to enumerate these, there are only $4$ of them, $(0,0,4),(0,1,3),(0,2,2),(1,1,2)$.  So, there are $24$ ways to allocate the balls in your problem.
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\begin{align}
?&amp;=\sum_{x_{1} = 1}^{9}\sum_{x_{2} = 1}^{9}\sum_{x_{3} = 1}^{9}
\delta_{x_{1} + x_{2} + x_{3}, 10}
=\sum_{x_{1} = 1}^{9}\sum_{x_{2} = 1}^{9}\sum_{x_{3} = 1}^{9}\int_{\verts{z} = 1}
{1 \over z^{11 - x_{1} - x_{2} - x_{3}}}\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=
\int_{\verts{z} = 1}\pars{\sum_{x = 1}^{9}z^{x}}^{3}{1 \over z^{11}}
\,{\dd z \over 2\pi\ic}
=
\int_{\verts{z} = 1}\bracks{z\pars{z^{9} - 1} \over z - 1}^{3}{1 \over z^{11}}
\,{\dd z \over 2\pi\ic}
=
\int_{\verts{z} = 1}{\pars{1 - z^{9}}^{3} \over \pars{1 - z}^{3}}{1 \over z^{8}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=
\int_{\verts{z} = 1}\sum_{\ell = 0}^{3}{3 \choose \ell}
\pars{-z^{9}}^{\ell}\sum_{\ell' = 0}^{\infty}{\ell' + 2 \choose 2}z^{\ell'}\,
{1 \over z^{8}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=
\sum_{\ell = 0}^{3}\sum_{\ell' = 0}^{\infty}\pars{-1}^{\ell}{3 \choose \ell}
{\ell' + 2 \choose 2}\delta_{9\ell + \ell',7}
=
\sum_{\ell = 0}^{3}\pars{-1}^{\ell}{3 \choose \ell}
\sum_{\ell' = 0}^{\infty}{\ell' + 2 \choose 2}\delta_{\ell',7 - 9\ell}
\\[3mm]&amp;=
\sum_{\ell\ =\ 0 \atop {\vphantom{\LARGE A}\ell\ \leq\ 7/9}}^{3}\pars{-1}^{\ell}
{3 \choose \ell}{9 - 9\ell \choose 2} = {9 \choose 2} = \color{#0000ff}{\Large 36}
\end{align}