# Quite Strange Commons, aren't they ?

How many non-negative integer solutions of the equation $a+2s+3d+4f+5g+6h=99$ are also the solutions of $q+w+o+r+t+y=99$

Details and assumptions :-

$$\bullet \quad$$ Solutions here means the oredered 6-tuples $$(a,s,d,f,g,h)$$ and $$(q,w,o,r,t,y)$$

Are also solutions of means the 6-tuple (a,s,d,f,g,h) which satisfies 1st equation, must also satisfy the 2nd equation (If $$(a_0,s_0,d_0,f_0,g_0,h_0)$$ is some solution tuple of 1st equation, then $$q=a_0 , w=s_0,e=d_0,r=f_0,t=g_0,y=h_0$$ must satisfy second equation.

$$\bullet \quad \quad a,s,d,f,g,h,q,w,o,r,t,y \in \mathbb{N} \cup$${0}. (All non-negative integers).

×