# Even stranger commons !!!

How many non-negative integer solutions of the equation $a+4b+c+3d+3e+3f+6g+6h=20$are also the solutions of $j+2k+l+3m+n+x+p+6q=20$

Details and assumptions:-

$$\bullet \quad$$ Solutions here means the ordered 8-tuples $$(a,b,c,d,e,f,g,h)$$ for the first equation and $$(j,k,l,m,n,x,p,q)$$ for the second equation.

$$\bullet \quad$$ Are also solutions of means the 8-tuple (a,b,c,d,e,f,g,h) which satisfies 1st equation, must also satisfy the 2nd equation.

(If $$a_0,b_0,c_0,d_0,e_0,f_0,g_0,h_0)$$ is some solution tuple of 1st equation, then $$j=a_0 , k=b_0,l=c_0,m=d_0,n=e_0,x=f_0,p=g_0,q=h_0$$ $$\color{Red}{\textbf{must}}$$ satisfy second equation.

$$\bullet \quad \quad a,b,c,d,e,f,g,h,j,k,l,m,n,x,p,q \in \mathbb{N} \cup$${0}. (All non-negative integers).

Problem 1 of this type was a bit easier, and it gives hint for this one.

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