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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.

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2,5,3? Those add up to ten and match your condition. Forgive me if I didnt understand your question right. –  Karim O. Jan 2 at 7:53
    
Never Mind. What we actually need is the number of possibilities that adds up to 10. –  suzy ong Jan 2 at 7:59
    
I guess [7,2,1] and [6,1,3] also work, not much combinations to work with –  Karim O. Jan 2 at 8:10

3 Answers 3

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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Assume for the moment that we want $x_1<x_2<x_3$. The we necessarily have $$\eqalign{x_1&=1+y_1,\cr x_2&=(x_1+1)+y_2=2+y_1+y_2,\cr x_3&=(x_2+1)+y_3=3+y_1+y_2+y_3\cr}$$ with $y_k\geq0$ $\>(1\leq k\leq 3)$. The condition $x_1+x_2+x_3=10$ then implies $$3y_1+2y_2+y_3=4\ ,\tag{1}$$ which automatically enforces $y_1+y_2+y_3\leq 6$, or $x_3\leq9$.

The nonnegative solutions of $(1)$ can be found by inspection: We obtain the triples $(0,0,4)$, $(0,1,2)$ and $(0,2,0)$ with $y_1=0$ and the single triple $(1,0,1)$ with $y_1=1$.

So there are $4$ solutions when the $x_k$ are in increasing order, or $24$ solutions when they can be in any order.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} ?&=\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]&= \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]&= \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]&= \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]&= \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}

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