# How many solutions has $x_1 + 2x_2 + 2x_3 = 200$ with non-negative integers?

Problem:
Let $$x_1$$, $$x_2$$ and $$x_3$$ be integers such that $$x_1 \geq 0$$, $$x_2 \geq 0$$ and $$x_3 \geq 0$$. How many solutions does the following equation have: $$x_1 + 2x_2 + 2x_3 = 200$$

Answer:
Let $$c$$ be the number of solutions of this equation. For the case where $$x_1$$ is odd, the equation has no solutions. Now I consider the smallest case of $$x_1$$ where $$x_1 = 0$$. I now have the following equation: $$2x_2 + 2x_3 = 200$$ or $$x_2 + x_3 = 100$$ This equation has $$101$$ solutions. Now I consider the case where $$x_0 = 2$$. I now have the following equation: $$2 + 2x_2 + 2x_3 = 200$$ or $$x_2 + x_3 = 99$$ This equation has $$100$$ solutions. Now I consider the case where $$x_0 = 4$$. I now have the following equation: $$4 + 2x_2 + 2x_3 = 200$$ or $$x_2 + x_3 = 98$$ This equation has $$99$$ solutions. Now I consider the case where $$x_0 = 200$$. I now have the following equation: $$100 + 2x_2 + 2x_3 = 200$$ or $$x_2 + x_3 = 0$$ This equation only has one solution.

\begin{align*} c &= \sum_{i = 0}^{100} i+1 = \sum_{i = 1}^{100} i + \sum_{i = 0}^{100} 1 \\ \sum_{i = 1}^{100} i &= \dfrac{ 100(101) }{2} = 50(101) \\ \sum_{i = 0}^{100} 1 &= 101 \\ c &= 50(101) + 101 \\ c &= 5151 \end{align*}

Is my solution correct?

## 2 Answers

Your solution seems correct, but a quicker way would be the following:

As you observed, $$x_1$$ has to be even, so the problem will have as many solutions as the number of solutions to the problem of

$$2x_1+2x_2+2x_3=200,$$

i.e.

$$x_1+x_2+x_3=100.$$

Now this problem is the same problem as the number of ways to split $$100$$ objects into $$3$$ containers, which can also be considered as the number of words with $$100$$ of one letter, and $$2$$ of another (think of the $$100$$ letters as your objects and the $$2$$ letters as sectioning them off into the containers). This problem is simple and gives you the number as

$$\frac{102!}{100!\cdot 2!}=\frac{101\cdot 102}{2}=101\cdot 51=5151.$$

• This is a nice solution. For clarity, however, I would suggest replacing $x_1$ with $2t$ (i.e., use some variable other than $x_1$). Mar 15, 2022 at 22:02

I do not know whether you are interested in a little bit advance but much more easier technique which is generating functions.Nonetheless , i thought that this may help you to get rid of long techniques and expand your vision over combinatorics problems. Anyway , here is my solution for you :

You say that $$x_1 \geq0 ,x_2 \geq 0 , x_3\geq 0$$ , so

• Possible values for $$x_1: 0,1,2,3,4,5,6,....$$

• Possible values for $$x_2: 0,1,2,3,4,5,6,....$$ . Then , the possible values for $$2x_2=0,2,4,6,8,....$$

• Possible values for $$x_3: 0,1,2,3,4,5,6,....$$ . Then , the possible values for $$2x_3=0,2,4,6,8,....$$

Now , write the generating functions ,which are polynomial series whose exponents satisfy the values that the variables can take, for each variable $$x_1,x_2,x_3$$ such that

• For $$x_1:$$ $$\frac{1}{1-x}=x^0+x^1+x^2+x^3 +x^4 +....$$

• For $$2x_2:$$ $$\frac{1}{1-x^2}=x^0+x^2+x^4+x^6 +x^8 +....$$

• For $$2x_3:$$ $$\frac{1}{1-x^2}=x^0+x^2+x^4+x^6 +x^8 +....$$

Now , if we find the product of these three series , we find a PRODUCT such that the product has exponent ,as well. Then ,the coefficient of the $$\color{blue}{x}$$ whose exponent is equal to $$\color{red}{200}$$ will give you the number of possible solutions satisfying $$x_1+x_2+x_3=200$$

$$\therefore$$ the answer is $$5151$$

$$\mathbf{\text{EDITION}:}$$ Calculation of result without using Wolfram-Alpha:

We need to find $$[x^{200}]\bigg(\frac{1}{1-x}\bigg)\bigg(\frac{1}{1-x^2}\bigg)^2$$

When you see generating function forms , $$[x^k]$$ means that we are looking for the coefficient of the term of $$x^k$$. Now , lets remember binomial expansion a little bit.

• The coefficient of the term $$x^k$$ in the expansion of $$(\frac{1}{1-x})$$ is equal to $$\binom{k+\color{red}{1}-1}{k}$$ where the red one represent the exponential of the $$(\frac{1}{1-x})$$ , so any term in this expansion is equal to $$\binom{k+\color{red}{1}-1}{k}x^k$$

• The coefficient of the term in the expansion of $$(\frac{1}{1-x^2})^2$$ is equal to $$\binom{m+\color{red}{2}-1}{k}$$ where the red two represent the exponential of the $$(\frac{1}{1-x^2})^{\color{red}{2}}$$ , so any term in this expansion is equal to $$\binom{m+\color{red}{2}-1}{m}(x^2)^m$$

In this expansion we look for $$2m+k=200$$ where

• $$m \in \{100,99,98,...,2,1,0\}$$

• $$k \in \{0,2,4,..,196,198,200\}$$

As you see the matching , for example , when $$m=100$$ then $$n=0$$ and when $$m=99$$ then $$n=2$$ so on. Then we need to find the summation of the coefficient of those cases such that

• When $$m=100$$ and $$n=0$$ , then the coefficient of $$x^{200}= \binom{100+2-1}{100}\binom{0+1-1}{0}$$

• When $$m=99$$ and $$n=2$$ , then the coefficient of $$x^{200}= \binom{99+2-1}{99}\binom{2+1-1}{2}$$

• When $$m=0$$ and $$n=200$$ , then the coefficient of $$x^{200}= \binom{0+2-1}{0}\binom{200+1-1}{200}$$

As you see , it is equal to $$101+100+...+1 = \frac{101 \times 102}{2}=5151$$

As you see my dear friend @trueblueanil , calculation of them without using software is not enjoyable in some cases..

• How do you actually get the value when Wolfram doesn't expand fully ? Mar 22, 2022 at 7:18
• @trueblueanil when you do not use wolfram , generally expanded binomial theorem is used for expansion.However , it is very long process for some types of questions. I am putting here some of my answer which i get in calculation . Link 1,,,,Link 2-) Mar 22, 2022 at 8:12
• @trueblueanil i wll edit the answer for you , wait please Mar 22, 2022 at 8:26
• Thanks for the added explanation. Yes, sometimes g.f's are easy to write down as an approach, but difficult to compute ! :-) Mar 22, 2022 at 8:57
• For the lucid added material, (+1) ! Mar 22, 2022 at 16:38