# Number of solutions

• Nov 15th 2010, 06:04 AM
kumaran5555
Number of solutions
Number of integral solutions of x+y+z=0 with x>=5 , Y>=-5 and z >=-5 ?

How to solve this by combination ?
• Nov 15th 2010, 06:42 AM
DrSteve
I assume that you mean integer solutions.

Off the top of my head I don't see how to do this with a single combination because different choices for x lead to different possibilities for y and z, and order seems to matter.

For example, if x = 5, then there are 6 possibilities for the pair (y,z): (-5, 0), (-4, -1), ..., (0,-5)
Similarly, if x = 6, then there are 5 possibilities for the pair (y,z).
Continuing in this fashion, we see that there are 6+5+4+3+2+1=21 possibilities for the triple (x,y,z).
• Nov 15th 2010, 06:50 AM
emakarov
Also, it's the same number as in y + z >= 5 where y <= 5 and z <= 5.
• Nov 15th 2010, 07:03 AM
kumaran5555
But i have answer as 136 !.

I don't think it is possible to have these many combination in the given range.

To have the equation, we need the sum of y+z to be negative.
• Nov 15th 2010, 07:12 AM
emakarov
Quote:

Originally Posted by kumaran5555
But i have answer as 136 !.

Between this thread and that, I am starting to think that your textbook (or whichever source of answers you have) is faulty.
• Nov 15th 2010, 07:39 AM
kumaran5555
I am really sorry to have that kind of material.

Apologies.
• Nov 15th 2010, 07:43 AM
aman_cc
Maybe this will work better -

Find coefficient of $\displaystyle x^0$ in $\displaystyle (x^5 + x^6 + .....)(x^{-5} + x^{-4} + .....)(x^{-5} + x^{-4} + .....)$
• Nov 15th 2010, 08:09 AM
Opalg
Quote:

Originally Posted by kumaran5555
But i have answer as 136 !.

I don't think it is possible to have these many combination in the given range.

To have the equation, we need the sum of y+z to be negative.

Have you copied the question correctly? If the problem is to find the number of integral solutions of x+y+z=0 with x>=5 , y>=–5 and z >=–5, then the answer is indeed 136, as you can check by using aman_cc's useful hint.
• Nov 15th 2010, 09:38 PM
kumaran5555
Thanks.

The problem statement says integral solutions only.

But can anyone tell me till how long i have to expand the series.

As per the question, i think there should be of some range.
• Nov 16th 2010, 02:53 AM
Plato
This link expands the sum for $\displaystyle x\ge 5$ which does not give 136.

This link expands the sum for $\displaystyle x\ge -5$ which does give 136.
• Nov 16th 2010, 06:44 AM
Soroban
Hello, kumaran5555!

Quote:

$\displaystyle \text{Number of integral solutions of: }\: x+y+z\:=\:0$
$\displaystyle \text{with }x \ge 5,\;y \ge -5,\;z \ge -5$

After approaching it five ways and getting five different answers,
. . I made a brute-force List.

The maximum value for $\displaystyle \,x$ is $\displaystyle x = 10.$

. . Then: .$\displaystyle (x,y,z)\:=\:(10,\text{-}5,\text{-}5)$

Then we have:
. . $\displaystyle \begin{array}{c} (10,\text{-}5,\text{-}5) \\ (9,\text{-}5,\text{-}4)\quad(9,\text{-}4,\text{-}5) \\ (8,\text{-}5,\text{-}3)\quad(8,\text{-}4,\text{-}4)\quad(8,\text{-}3,\text{-}5) \\ (7,\text{-}5,\text{-}2)\quad(7,\text{-}4,\text{-}3)\quad(7,\text{-}3,\text{-}4)\quad(7,\text{-}2,\text{-}5) \\ (6,\text{-}5,\text{-}1)\quad(6,\text{-}4,\text{-}2)\quad(6,\text{-}3,\text{-}3)\quad(6,\text{-}2,\text{-}4)\quad(6,\text{-}1,\text{-}5) \\ (5,\text{-}5,0)\quad(5,\text{-}4,\text{-}1)\quad(5,\text{-}3,\text{-}2)\quad(5,\text{-}2,\text{-}3)\quad(5,\text{-}1,\text{-}4)\quad(5,0,\text{-}5) \end{array}$

I found 21 solutions.

• Nov 16th 2010, 07:15 AM
kumaran5555