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Question:

What is the number of solutions to the equation:

$x_1+x_2+\cdots+x_m=n$

when $\forall \ i : 1\le x_i\le n$

My approach:

I took $\ $ $\forall \ i : y_i= x_i-1 \ $ and then i got the equation:

$\ y_1+y_2+\cdots+y_m=n-m \ $ when $\ \forall \ i : 0\le y_i\le n-1$.

And now I take $\ $ $\forall \ i : z_i= n-1-y_i \ $ so now the equation will be:

$\ z_1+z_2+\cdots+z_m=n-1-y_1+\cdots+n-1-y_m=m(n-1)-(y_1+\cdots+y_m)=m(n-1)-(n-m)=mn-m-n+m=n(m-1) \ $

to make a long story short:

$\ z_1+\cdots+z_m=n(m-1) \ $ when $\ \forall \ i : 0\le z_i$

and for such an equation the number of solutions is simply $\binom{(n+1)(m-1)}{n(m-1)}$

please help me. Did I make some mistakes or is it a good solution.

Community
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Don Fanucci
  • 2,485

2 Answers2

3

Your solution is incorrect.

Method 1: Solving the problem in the non-negative integers.

We wish to find the number of solutions to the equation $$x_1 + x_2 + \cdots + x_m = n \tag{1}$$ where each $x_k$, $1 \leq k \leq m$, is a positive integer. As you observed, if we set $y_k = x_k - 1$, then $y_k$ is a non-negative integer and $$y_1 + y_2 + \cdots + y_m = n - m \tag{2}$$ which has the same number of solutions in the non-negative integers that equation 1 has in the positive integers. A particular solution of equation 2 corresponds to the placement of $m - 1$ addition signs in a row of $n - m$ ones.

For example, if $n = 10$ and $m = 3$, then $n - m = 7$, so $$1 1 + + 1 1 1 1 1$$ corresponds to the solution $y_1 = 2$, $y_2 = 0$, and $y_3 = 5$, while $$1 1 1 1 + 1 1 + 1$$ corresponds to the solution $y_1 = 4$, $y_2 = 2$, and $y_3 = 1$.

Thus, the number of solutions of equation 2 is the number of ways we can select which $m - 1$ of the $(n - m) + (m - 1) = n - 1$ symbols ($n - m$ ones and $m - 1$ addition signs) are addition signs, which is $$\binom{n - 1}{m - 1}$$

Method 2: Solving the problem in the positive integers.

A particular solution of equation 1 in the positive integers corresponds to the placement of $m - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones. Hence, the number of solutions of equation in the positive integers is $$\binom{n - 1}{m - 1}$$

For example, if $n = 10$ and $m = 3$. Then the solution $$1 1 1 + 1 + 1 1 1 1 1 1$$ corresponds to $x_1 = 3$, $x_2 = 1$, and $x_3 = 6$, while the solution $$1 1 1 1 1 + 1 1 1 + 1 1$$ corresponds to $x_1 = 5$, $x_2 = 3$, and $x_3 = 2$. Notice that these cases correspond to the two examples given above when we solved the problem in the non-negative integers.

N. F. Taussig
  • 76,571
2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\quad\pars{~x_{i} \in \mathbb{N}_{\ \geq\ 1}\,,\quad i = 1, 2, \ldots, m~},\quad}$ the number of configurations is given by:

\begin{align} &\sum_{x_{1}\ =\ 1}^{\infty}\cdots\sum_{x_{m}\ =\ 1}^{\infty} \bracks{x_{1} + \cdots + x_{m} = n} = \sum_{x_{1}\ =\ 1}^{\infty}\cdots\sum_{x_{m}\ =\ 1}^{\infty}\,\,\, \oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n + 1 - x_{1} - \cdots - x_{m}}}\, \,\,\,\,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n + 1}} \pars{\sum_{x = 1}^{\infty}z^{x}}^{m}\,{\dd z \over 2\pi\ic} = \oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n + 1}}\,\pars{z \over 1 - z}^{m} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n - m + 1}}\,\,\pars{1 - z}^{-m} \,\,\,{\dd z \over 2\pi\ic} = \bracks{z^{n - m}}\pars{1 - z}^{-m} \\[5mm] = &\ {-m \choose n - m}\pars{-1}^{n - m} = \braces{{m +\bracks{n - m} - 1\choose n - m}\pars{-1}^{\, n - m}} \pars{-1}^{\, n - m} = \bbx{\ds{n - 1 \choose n - m}} \end{align}

Felix Marin
  • 89,464