0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Barry Cipra Nov 6 '18 at 16:42
  • $\begingroup$ @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, $\sum_{i=0}^6 {3+i\choose i}= 210$. $\endgroup$ – irchans Nov 6 '18 at 17:01
  • $\begingroup$ 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.) $\endgroup$ – Barry Cipra Nov 6 '18 at 17:47
  • $\begingroup$ 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($x_1+x_2+x_3 \leq11 )$ $\endgroup$ – user3767495 Nov 7 '18 at 3:20
0
$\begingroup$

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 ;)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.