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 $1?$

cooIcooIcooI17 Oct 10, 2024

#1**0 **

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**

Maxematics Oct 11, 2024