Question

Find the number of positive unequal integral solution of the equation $x+y+z+w=20$

Answer 1

We can assume that $x<y<z<w$ without loss of generality.
Not put $x_1=x,x_2=y-x,x_3=z-y$ and $x_4=w-z$
Then $x_1,x_2,x_3,x_4>1$
$x=x_1,{y=x}_1+x_2,z=x_1{+x}_2+x_3$ and $w=x_1+x_2+x_3+x_4$
and the given equation becomes ${4x}_1+{3x}_2+{2x}_3+x_4=20$
The number of positive integral solution of this equation equal to the coefficient of $t^{29}$ in
$(t^4+t^8+t^{12}+...)(t^3+t^6+t^9+...)(t^2+t^4+t^6+...)(t+t^2+t^3+...)=t^{10}(1+t^4+t^8+...)(t+t^3+t^6+...)(1+t^2+t^4+...)(1+t+t^2+...)$
i.e the coefficient of $t^{10}$ in
$(1+t^4+t^6)(1+t^3+t^6+t^9)(1+t^2+...+t^{10})(1+t+t^2...t^{10})$
This is the coefficient of $t^{10}$ in
$(1+t^3+t^4+t^6+t^7+t^8+t^9+t^{10})\times (1+t+{2t}^2+{2t}^3+{3t}^4+{3t}^5+{4t}^6+{4t}^7+{5t}^2+{5t}^9+{6t}^{10})$
which is $6+4+4+3+2+2+1+1=23$
Furthermore since $x,y,z$ and $w$ can be permuted in ${}^4{P}_4=4!=24$ ways
The required number of solution is $\left(23\right)\left(24\right)=552$