# Number of integral solutions to the equation

How many positive integral solutions to the below equation?

$$x_1+x_2+x_3+x_4 \leq10$$

I tried like below

Since we want positive integral solutions so it means

$$x_1 \geq1, x_2 \geq 1, x_3 \geq 1, x_4 \geq 1$$

so inequality transforms to

$$x_1+x_2+x_3+x_4 \leq 6$$ with conditions $$x_1,x_2,x_3,x_4 \geq 0$$

Now I break this into 7 parts and add result

$$x_1+x_2+x_3+x_4=0\Rightarrow \binom{3}{0} ways$$

$$x_1+x_2+x_3+x_4=1 \Rightarrow \binom{4}{1} ways$$

$$x_1+x_2+x_3+x_4=2 \Rightarrow \binom{5}{2} ways$$

$$x_1+x_2+x_3+x_4=3 \Rightarrow \binom{6}{3} ways$$

$$x_1+x_2+x_3+x_4=4 \Rightarrow \binom{7}{4} ways$$

$$x_1+x_2+x_3+x_4=5 \Rightarrow \binom{8}{5} ways$$

$$x_1+x_2+x_3+x_4=6 \Rightarrow \binom{9}{6} ways$$

But I got wrong answer. Where I went wrong?

• Most of your other questions so far have included some description of what you've tried and where you got stuck. It would help if you did so here as well. – Barry Cipra Nov 6 '18 at 16:42
• @user3767 As far as I can tell, your approach is correct. Let $$n=|\{(x_1, x_2, x_3, x_4) | \sum x_i \leq 10, x_i\geq 1, x_i\in \mathbb N\}|.$$ It's pretty easy to write a computer program which gives the answer $n=210$. And using your approach, $\sum_{i=0}^6 {3+i\choose i}= 210$. – irchans Nov 6 '18 at 17:01
• What (final, numerical) answer did you get, and why do you think it's wrong? (BTW, thanks for your edit in response, I assume, to my first comment.) – Barry Cipra Nov 6 '18 at 17:47
• The final answer suggested that I should take another variable $x_5$ such that $x_1+x_2+x_3+x_4+x_5=10$. And then it gives $\binom{10+5-1}{10}ways$ Even in Kenneth Rosen-Indian Adaptation Edition-7-Page379 Problem Number 20, is similar to it($x_1+x_2+x_3 \leq11 )$ – user3767495 Nov 7 '18 at 3:20

Since the solutions have to be positive, we can make this substitution:

$$x_i=1+y_i \ \ \ \ i=1,2,3,4$$

The disequation becomes:

$$y_1+y_2+y_3+y_4\leq 6$$

And here is the trick. Imagine to give some of 6 candies to 4 children, the others one are thrown in the trash can. So let's create a "trash can" variable $$y_5$$:

$$y_1+y_2+y_3+y_4+y_5=6$$

You should know how to solve this ;)