We have,  $x+y+z+w=20 ...\left(i\right)$

Assume $x<y<z<w.$ Here, $x,y,z,w\ge 1$

Now, let $x=x_1,y-x=x_2,z-y=x_3$ and $w-z=x_4$

$\begin{array}{l}\therefore x=x_1,y=x_1+x_2, z=x_1+x_2+x_3\text{ and }\\ w=x_1+x_2+x_3+x_4\end{array}$

From Eq. (i), $4x_1+3x_2+2x_3+x_4=20$

Then, $x_1,x_2,x_3,x_4\ge 1$

$\because 4x_1+3x_2+2x_3+x_4=20 ...\left(ii\right)$

$\therefore$ Number of positive integral solutions of Eq. (ii)

   $=$ Coefficient of $x^{20-10}$ in ${\left(1-x^4\right)}^{-1}{\left(1-x^3\right)}^{-1}{\left(1-x^2\right)}^{-1}(1-x{)}^{-1}$

  $=$Coefficient of $x^{10}$ in ${\left(1-x^4\right)}^{-1}{\left(1-x^3\right)}^{-1}{\left(1-x^2\right)}^{-1}(1-x{)}^{-1}$

  $=$Coefficient of $x^{10}$ in $\left(1+x^4+x^8+x^{12}+\dots \right)\times \left(1+x^3+x^6+x^9+x^{12}+\dots \right)\times \left(1+x^2+x^4+x^6+x^8+x^{10}+\dots \right)\times \left(1+x+x^2+x^3+x^4\right.\left.+x^5+x^6+x^7+x^8+x^9+x^{10}+\dots \right)$

  $=$Coefficient of $x^{10}$ in $\left(1+x^3+x^6+x^9+x^4+x^7+x^{10}+x^8\right)x\left(1+x^2+x^4+x^6+x^8+x^{10}\right)\left(1+x+x^2+x^3\left.+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}\right)\right.$[neglecting higher powers] 

  $=$Coefficient of $x^{10}$ in $\left(1+x^2+x^4+x^6+x^8+x^{10}+x^3+x^5+x^7+x^9+x^6\right.+x^8+x^{10}+x^9+x^4+x^6+x^8+x^{10}+x^7+x^9+x^{10}\left.+x^8+x^{10}\right)\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7\right.\left.+x^8+x^9+x^{10}\right)$

  [neglecting higher powers] 

$=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1=23$

But $x, y, z$ and $w$ can be arranged in ${}^{4}P_4=4!=24$

Hence, required number of Sols $s=\left(23\right)\left(24\right)=552$