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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 $2?$

 May 8, 2023
 #1
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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​.

 May 8, 2023
 #2
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[12 C 4]  +  [11 C 4]  +  [10 C 4] ==495  +  330  + 210==1,035 - total number of solutions.

 May 8, 2023

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