# Quite Strange Commons, aren't they ?

**Discrete Mathematics**Level 3

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).

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