How many 3-tuples satisfy $x_{1} + x_{2} + x_{3} = 11;$ $(x_{1} ,x_{2} ,x_{3}$ are nonnegative integers?)

Question

I know that the total number of choosing without constraint is

$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13·12}{2} =78$

Then with x1 ≥ 1, x2 ≥ 2, and x3 ≥ 3.

the textbook has the following solution

$\binom{3+5−1}{5}=\binom{7}{5}=21$ I can't figure out where is the 5 coming from?

The reason to choose 5 is because the constraint adds up to 6? so 11 -6 =5?

Answer 1

For the given equation: $$x_1+x_2+x_3 = 11, \;\text{ with } x_1, x_2, x_3 \;\text{ non-negative },$$ your solution is correct.

$$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13!}{2!11!} = \frac{13·12}{2} =78$$

Your final answer is correct, (Now corrected: (but you failed to show that you need to divide $13\times 12$ by $2$ to obtain $78$).

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The other solution (from the text) would be the solution to $x_1+x_2+x_3 = 5$, with $x_1, x_2, x_3$ non-negative. In this case, there are $$\binom{3+5-1}{5} = \frac{7!}{2!5!} = \frac{7\cdot 6}{2} = 21$$

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Edit after another question-update: We now are solving $$x_1 + x_2 + x_3 = 11, \text{ with }\; x_1\geq 0+1,\,x_2 \geq 0 + 2,\,x_3 \geq 0 + 3.$$

We can solve in the same manner by writing $(x_1+1)+ (x_2 + 2) + (x_3+3) = 11-1-2-3 = 5$.

Then we may simply ascribe $y = x_1 + 1, y_2 = x_2+2, y_3 = x_3 + 3$ to get $$y_1+y_2 +y_3 = 5$$

From here, the text's solution (addressed above) solves the number of the required solutions.

Answer 2

The number of nonnegative integer solutions of $x_1 + x_2 + x_3 = 11$ is the coefficient of $t^{11}$ in the following generating function [JDL]

$$\dfrac{1}{(1-t)^3}$$

Suppose now that we are interested in integer solutions with $x_1 \geq 1$, $x_2 \geq 2$ and $x_3 \geq 3$. We thus introduce three new nonnegative variables

$$z_1 : = x_1 - 1 \qquad\qquad\qquad z_2 : = x_2 - 2 \qquad\qquad\qquad z_3 : = x_2 - 3$$

The number of admissible integer solutions of $x_1 + x_2 + x_3 = 11$ is the number of nonnegative integer solutions of $z_1 + z_2 + z_3 = 5$, which is the coefficient of $t^5$ in the generating function.

Using SymPy:

>>> from sympy import *
>>> t = Symbol('t')
>>> f = 1 / (1-t)**3
>>> f.series(t,0,12)
1 + 3*t + 6*t**2 + 10*t**3 + 15*t**4 + 21*t**5 + 28*t**6 + 36*t**7 + 45*t**8 + 55*t**9 + 66*t**10 + 78*t**11 + O(t**12)

Hence, the number of admissible integer solutions is $21$. Note that the coefficients in the series are the triangular numbers (A000217)

$$\binom{2}{2}, \binom{3}{2}, \binom{4}{2}, \binom{5}{2}, \dots, \binom{k+2}{2}, \dots$$

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[JDL] Jesús A. De Loera, The Many Aspects of Counting Lattice Points in Polytopes.

Answer 3

$$(1,0,10),(2,3,6),(1,2,8),(1,3,7)$$ which satisfies above equation, if rearranged gives 24 solutions and you have even more solutions. So your text book answer is wrong and I dont see any problems in ur method.

Answer 4

This can be solved also using the stars and bars method. The point is paying attention to variables that take value 0. So you have 3 cases:

1) all variables $\ne 0$ This amounts to $\binom{11-1}{3-1}=45$

2) just one variable have value $0$ (and hence two others are $\ne 0$) This amounts to $\binom{3}{1}\cdot\binom{11-1}{2-1}=30$

3) two variables are $0$ (and hence only one is non-zero This amounts to $\binom{3}{2}\cdot 1=3$

Taking 1)+2)+3) gives 78

as already found with the other methods.

For the second part, you just have to adjust your question to the new constraints, that is to say $x_1+x_2+x_3=8$ (can you see this?). Applying again the stars and bars method you find

$\binom{8-1}{3-1}=21$

The situation is simpler in this second part since all variables are $\ne 0$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \bracks{z^{11}}\sum_{x_{1} = 1}^{\infty}z^{x_{1}} \sum_{x_{2} = 2}^{\infty}z^{x_{2}}\sum_{x_{3} = 3}^{\infty}z^{x_{3}} & = \bracks{z^{11}}{z \over 1 - z}\,{z^{2} \over 1 - z}\,{z^{3} \over 1 - z} = \bracks{z^{\color{#f00}{5}}}\pars{1 - z}^{-3} \\[5mm] & = \bracks{z^{5}}\sum_{i = 0}^{\infty}{-3 \choose i}\pars{-z}^{i} = -{-3 \choose 5} = {7 \choose 5} = {7 \times 6 \over 2} = \bbx{\ds{21}} \end{align}

Note that $\ds{\color{#f00}{5} = 11 - 1 - 2 - 3}$.