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# counting question

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How many solutions are there to the equation

\(u + v + w + x + y + z = 8\)
where \(u, v, x, y, \) and \(z\) are nonnegative integers, and \(x \) is at most 2?

Apr 16, 2023

### 2+0 Answers

#1
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Since x is at most 2, we have x∈0,1,2. We can break down the problem into three cases based on the value of x:

Case 1: x=0.

If x=0, then we need to find the number of solutions to the equation u+v+w+y+z=8, where u,v,w,y,z are nonnegative integers. This is a standard stars and bars problem, and the number of solutions is C(8−1+2,2​)=36.

Case 2: x=1.

If x=1, then we need to find the number of solutions to the equation u+v+w+1+y+z=8, where u,v,w,y,z are nonnegative integers. This is also a stars and bars problem, and the number of solutions is C(8−1+2,2​)=36.

Case 3: x=2.

If x=2, then we need to find the number of solutions to the equation u+v+w+2+y+z=8, where u,v,w,y,z are nonnegative integers. This is also a stars and bars problem, and the number of solutions is C(8−1+2,2)=36.

Therefore, the total number of solutions is 36+36+36=108​.

Apr 16, 2023
#2
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a=listfor(n, 1, 3, (9+n) nCr 4);print a, "=", sum a

(210, 330, 495) = 1035 - total number of solutions.

Apr 17, 2023