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

How to solve this by combination ?

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- Nov 15th 2010, 06:04 AMkumaran5555Number 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 AMDrSteve
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 AMemakarov
Also, it's the same number as in y + z >= 5 where y <= 5 and z <= 5.

- Nov 15th 2010, 07:03 AMkumaran5555
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 AMemakarov
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 AMkumaran5555
I am really sorry to have that kind of material.

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

Find coefficient of in - Nov 15th 2010, 08:09 AMOpalg
- Nov 15th 2010, 09:38 PMkumaran5555
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 AMPlato
This link expands the sum for which does not give 136.

This link expands the sum for which does give 136. - Nov 16th 2010, 06:44 AMSoroban
Hello, kumaran5555!

Quote:

After approaching it five ways and getting five different answers,

. . I made a brute-force List.

The maximum value for is

. . Then: .

Then we have:

. .

I found**21**solutions.

- Nov 16th 2010, 07:15 AMkumaran5555
Thanks for your answer.