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.

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.
