+570 How many solutions are there to the equation
u + v + w + x + y + z = 2,
where $u,$ $v,$ $w,$ $x,$ $y,$ and $z$ are nonnegative integers, and $x$ is at most $1?$
+135 Case 1: x = 1
Then u + v + w + y + z = 1, and since none of them can be negative, exactly one of (u, v, w, y, z) is one, and the rest are zero. There are 5 ways to do this.
Case 2: x = 2
Then u + v + w + y + z, and using the same logic, the only case is u = v = w = y = z = 0.
Therefore there are 5 + 1 = 6 solutions
