Number of solutions

**kumaran5555** · November 15th 2010, 06:04 AM

Number of integral solutions of x+y+z=0 with x>=5 , Y>=-5 and z >=-5 ?

How to solve this by combination ?

**DrSteve** · November 15th 2010, 06:42 AM

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).

**emakarov** · November 15th 2010, 06:50 AM

Also, it's the same number as in y + z >= 5 where y <= 5 and z <= 5.

**kumaran5555** · November 15th 2010, 07:03 AM

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.

**emakarov** · November 15th 2010, 07:12 AM

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.

**kumaran5555** · November 15th 2010, 07:39 AM

I am really sorry to have that kind of material.

Apologies.

**aman_cc** · November 15th 2010, 07:43 AM

Maybe this will work better -

Find coefficient of $x^0$ in $(x^5 + x^6 + .....)(x^{-5} + x^{-4} + .....)(x^{-5} + x^{-4} + .....)$

**Opalg** · November 15th 2010, 08:09 AM

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.

**kumaran5555** · November 15th 2010, 09:38 PM

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.

**Plato** · November 16th 2010, 02:53 AM

This link expands the sum for $x\ge 5$ which does not give 136.

This link expands the sum for $x\ge -5$ which does give 136.

**Soroban** · November 16th 2010, 06:44 AM

Hello, kumaran5555!

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

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

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

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

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

I found 21 solutions.

**kumaran5555** · November 16th 2010, 07:15 AM

Thanks for your answer.

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