How many solutions are there to the equation \[u + v + w + x + y + z = 8,\]where $u,$ $v,$ $w,$ $x,$ $y,$ and $z$ are nonnegative integers, and $x$ is at most $2?$

Guest May 8, 2023

#1**0 **

The problem can be broken down into 3 cases:

Case 1: x=0

In this case, we need to find the number of solutions to the equation u+v+w=8, where u,v,w are nonnegative integers. This is a standard stars and bars problem, and the number of solutions is c(10,2)=45.

Case 2: x=1

In this case, we need to find the number of solutions to the equation u+v+w+x=8, where u,v,w,x are nonnegative integers. This is also a standard stars and bars problem, and the number of solutions is c(11,3) =165.

Case 3: x=2

In this case, we need to find the number of solutions to the equation u+v+w+x=8, where u,v,w,x are nonnegative integers. This is also a standard stars and bars problem, and the number of solutions is c(12,4)=495.

Therefore, the total number of solutions is 45+165+495=705.

Guest May 8, 2023