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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 $1?$

 Jun 21, 2024
 #1
avatar+129895 
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Case 1

x = 0  and  one  of the other variables = 2  →  5 different solutions

 

Case 2

x = 0   and two of the other variables = 1  →  C(5,2)  = 10 different solutions

 

Case 3

x = 1   and one  of the other variables = 1 →  C(5,1)  = 5 different solutions

 

5 + 10 + 5  =     20 possible solutions

 

cool cool cool

 Jun 21, 2024

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