Number of integral solutions of x+y+z=0 with x>=5 , Y>=-5 and z >=-5 ?
How to solve this by combination ?
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).
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.
Hello, kumaran5555!
$\displaystyle \text{Number of integral solutions of: }\: x+y+z\:=\:0$
$\displaystyle \text{with }x \ge 5,\;y \ge -5,\;z \ge -5$
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.