<s>[INST] ordered pair of unequal positive integer solution of $x+y+z+w = 20$ [1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$\bf{My\; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,z\in \mathbb{Z^{+}}$
$\bullet $ If $x=1$, Then $y+z=9$, So $(y,z) = (2,7)\;,(3,6)\;,(4,5)\;(5,4)\;,(6,3)\;,(7,2)$
$\bullet $ If $x=2$ Then $y+z=8$, So $(y,z) = (1,7)\;,(3,5)\;,(5,3)\;(7,1)$
$\bullet $ If $x=3$ Then $y+z=7$, So $(y,z) = (1,6)\;,(2,5)\;,(5,2)\;(6,1)$
$\bullet $ If $x=4$ Then $y+z=6$, So $(y,z) = (1,5)\;,(5,1)$.
$\bullet $ If $x=5$ Then $y+z=5$, So $(y,z) = (1,4)\;,(2,3)\;,(3,2)\;(4,1)$
$\bullet $ If $x=6$ Then $y+z=4$, So $(y,z) = (1,3)\;,(3,1)$
$\bullet $ If $x=7$ Then $y+z=3$, So $(y,z) = (1,2)\;,(2,1)$
So Total unordered pair is $ = 24$
My Question is , is there is any other Method to calculate the ordered pair in less complex way
because above is very Lengthy method
Help Required
Thanks.
 [/INST] For the first question, let's look at this for a general sum, $S$.  We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$.  We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have 
$$
x+(x+1)+(x+2) \le S
$$
so $3x+3 \le S$, i.e., $x\le (S/3)-1$.
As well, $y+(y+1)\le S-x$, i.e., $2y+1 \le S-x$, i.e. $y \le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=\sum_{x:1 \le x \le \frac{S}{3}-1} \sum_{y:x+1\le y \le \frac{S-x-1}{2}} 1
= \sum_{x:1 \le x \le \frac{S}{3}-1} \left( \left\lfloor \frac{S-x-1}{2}\right\rfloor -x\right)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=\mathrm{round}\left( \frac{(S-3)^2}{12} \right)$$
and I've verified that for $6\le S\le 10000$, so it's pretty surely correct.  A proof is probably among the many references at that OEIS entry.
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