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What is the number of solutions for the inequality $$x_1+x_2+x_3 \leq 11$$ where $x_1$, $x_2$, $x_3$ are integers greater than $0$?

Answer options: 182, 364 (correct), 78, 440

original problem image

I tried doing this using combination method

Since $x_1, x_2, x_3$ are integers greater than $0$, therefore value of $x_1 + x_2 + x_3$ satisfies $3 \leq x_1+x_2+x_3 \leq 11$.

$$_2C_2 \;+\; _3C_2 \;+\; _4C_2 \;+\; \cdots \;+\; _{10}C_2$$

And my answer was $165$ but that is not correct. I am not able to figure out where I am making a mistake

Jam
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Piyush Yadav
  • 191

3 Answers3

3

Your answer is correct for the stated problem. The given answer would be correct if $x_1, x_2, x_3$ were required to be nonnegative integers.

Here is another approach.

Since $x_1, x_2, x_3$ are positive integers, $x_1' = x_1 - 1$, $x_2' = x_2 - 1$, and $x_3' = x_3 - 1$ are nonnegative integers. Substituting $x_1' + 1$ for $x_1$, $x_2' + 1$ for $x_2$, and $x_3' + 1$ for $x_3$ in the inequality $$x_1 + x_2 + x_3 \leq 11 \tag{1}$$ yields \begin{align*} x_1' + 1 + x_2' + 1 + x_3 + 1 \leq 11\\ x_1' + x_2' + x_3' \leq 8 \end{align*} Let the slack variable $s = 8 - (x_1' + x_2' + x_3')$. Then $s$ is also a nonnegative integer. Moreover, $$x_1' + x_2' + x_3' + s = 8 \tag{2}$$ Thus, the number of solutions of inequality 1 in the positive integers is equal to the number of solutions of equation 2 in the nonnegative integers. A particular solution of equation 2 in the nonnegative integers corresponds to the placement of $4 - 1 = 3$ addition signs in a row of eight ones. For instance, $$1 1 + + 1 1 1 1 + 1 1$$ corresponds to the solution $x_1' = 2$, $x_2' = 0$, $x_3' = 4$, $s = 2$ ($x_1 = 3$, $x_2 = 1$, $x_3 = 5$). The number of such solutions is the number of ways we can place three addition signs in a row of eight ones, which is $$\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = 165$$ since we must choose which three of the eleven positions required for eight ones and three addition signs will be filled with addition signs.

Your answer $$\sum_{j = 2}^{10} \binom{j}{2} = \binom{11}{3}$$ by the hockey-stick identity, which states that $$\sum_{j = k}^{n} \binom{j}{k} = \binom{n + 1}{k + 1}$$ To see why it holds, we count $(k + 1)$-element subsets of the $(n + 1)$-element set $S = \{0, 1, 2, \ldots, n\}$ in two different ways. There are $$\binom{n + 1}{k + 1}$$ such subsets.

The number of such subsets with largest element $j$ is $$\binom{j}{k}$$
since each such subset must contain exactly $k$ of the $j$ elements in $S$ smaller than $j$. Since the $(k + 1)$-element subset of $S$ contains $k$ of the $j$ elements of $S$ less than $j$, $k \leq j \leq n$. Hence, the number of such subsets is $$\sum_{j = 0}^{n} \binom{j}{k}$$

Since we have counted the same $(k + 1)$-element subsets of the $(n + 1)$-element set $S$ in two different ways, the hockey-stick identity follows.

How was the stated answer obtained?

Finally, note that the stated answer of $364$ would be correct if the problem required $x_1, x_2, x_3$ to be greater than or equal to zero, in which case we would be solving the inequality in the nonnegative integers. If we let $$s = 11 - (x_1 + x_2 + x_3)$$ then the number of solutions of inequality 1 in the nonnegative integers is equal to the number of solutions of the equation $$x_1 + x_2 + x_3 + s = 11 \tag{3}$$ in the nonnegative integers. Equation 3 has $$\binom{11 + 4 - 1}{4 - 1} = \binom{14}{3} = 364$$ solutions in the nonnegative integers.

N. F. Taussig
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Using the method of generating functions (for more details, please see this asnwer and this one, I hope both are pretty detailed).

The number of solutions for $x+y+z=n, x\geq1, y\geq1, z\geq1$ is the coefficient of $x^n$ term of the $$(x+x^2+x^3+...+x^k+...)^3=\color{red}{\frac{x^3}{(1-x)^3}}$$ which is $\color{green}{\frac{(n-1)(n-2)}{2}=\binom{n-1}{2}}$. The final answer is $$\sum\limits_{n=3}^{11}\binom{n-1}{2}=165$$

because for $n \in \{1,2\}$ we have $\frac{(n-1)(n-2)}{2}=0$. So, you are right.

The number of solutions for $x+y+z=n, x\geq0, y\geq0, z\geq0$ is the coefficient of $x^n$ term of the $$(1+x+x^2+...+x^k+...)^3=\color{red}{\frac{1}{(1-x)^3}}$$ which is $\color{green}{\frac{(n+1)(n+2)}{2}=\binom{n+2}{2}}$. The final answer is $$\sum\limits_{n=0}^{11}\binom{n+2}{2}=364$$

Now, that you have two different answers with nearly different methods and looking at the image you posted, you can use this thread to complain.

rtybase
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Perhaps the shortest proof of the fact that we can choose $3$ positive integers summing to $\le 11$ in $\binom{11}{3}$ ways is that choosing the $x_i>0$ is equivalent to choosing distinct values from $1$ to $11$ of $\sum_{i=1}^jx_i$ for $1\le j\le 3$. For example, $x_1=2,\,x_2=3,\,x_3=2$ is equivalent to partial sums $2,\,5,\,7$. The case with non-negative integers can be incremented to replace the maximum of $11$ with $14$.

J.G.
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