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 $1?$

eramsby1O1O May 26, 2024

#1**0 **

To find the number of solutions to the equation u + v + w + x + y + z = 2, where u, v, w, x, y, z are nonnegative integers and x is at most 1, we can break it down into cases based on the value of x:

1. If x = 0: In this case, the equation becomes u + v + w + y + z = 2. Since the sum of the nonnegative integers is 2, there are 6 choose 4 = 15 solutions to this equation.

2. If x = 1: In this case, the equation becomes u + v + w + y + z = 1. Similarly, there are 5 choose 4 = 5 solutions to this equation.

Therefore, the total number of solutions to the given equation is 15 + 5 = 20.

Hi69420 May 26, 2024