Integer solutions of a less than inequality

Question

I need to determine how many integer solutions are to this inequality:$$ y_1 +y_2 +y_3 < 100 $$

with $$y_1 > 0,y_2 ∈ [0,10],y_3 ∈ (0,19]$$

I'm having trouble where to start. I know that$$y_1+y_2+y_3=99 $$since $100 $ is not possible to reach. Then I get stuck upon what I should do next.

The sum can be any integer in $[2, 99]$. – Macavity Jun 29 '15 at 15:33 — Macavity, Jun 29 '15 at 15:33

score 4 · Answer 1 · answered Jun 29 '15 at 15:35

For any given values of $y_2$ and $y_3$, there are $99 - y_2 - y_3$ values of $y_1$ that work.

Therefore, the answer is $$ \sum_{i = 0}^{10} \sum_{j=1}^{19} 99 - i - j $$

Which you can calculate using the formula $1+2+ \dots n = \dfrac{n(n+1)}{2}$.

user84413 · Answer 2 · 2015-06-29T19:10:02.573

1

Another method would be to let $y_4=99-(y_1+y_2+y_3)$, $\;\;$so $\;\;y_1+y_2+y_3+y_4=99$.

Now let $t_1=y_1-1, \;\;t_2=y_2, \;\;t_3=y_3-1, \;\;t_4=y_4\;\;$ so that $t_i\ge0$ for each i.

Then you want to find the number of integer solutions to

$\;\;\;\;t_1+t_2+t_3+t_4=97\;\;$ where $\;\;t_1\ge0, \;\;0\le t_2\le10,\;\; 0\le t_3\le18, \;\;t_4\ge0$,

which gives $\displaystyle\binom{100}{3}-\binom{89}{3}-\binom{81}{3}+\binom{70}{3}=17,556$.

edited Jun 29 '15 at 19:10

answered Jun 29 '15 at 17:25

user84413

Could you explain the values of t1, t2, t3, and t4 of why did you do that (like the reasoning behind it)? – Miriam Jun 29 '15 at 18:48
Sure, I will edit my answer to try to explain this. – user84413 Jun 29 '15 at 19:05
I thought that y1+y2+y3+y4=100 since a restriction of y4 is including 0 to infinity? – Miriam Jun 29 '15 at 19:13
We wanted $y_1+y_2+y_3\le99$, so we can let $y_4=99-(y_1+y_2+y_3)$. – user84413 Jun 29 '15 at 19:48
That kind of makes sense. I thought that adding an extra variable gives the inequality a placeholder. Also, could you explain the way you figured out the combinations for the solution? My book doesn't explain it very well. – Miriam Jun 29 '15 at 20:03
1

@Miriam: The extra variable takes up the slack between the sum of the three ‘real’ ones and the maximum possible sum of $99$. For the other part, take a look at my answer to this question and the one linked from that answer, and see if they help. – Brian M. Scott Jun 29 '15 at 20:44
why is it C(81,3)? – Miriam Jun 30 '15 at 01:38
I did y3>=20 and got C(80,3). – Miriam Jun 30 '15 at 02:09

2 Answers2