How many solutions are there to the equation
u + v + w + x + y + z = 10
where u, v, w, x, y, and z are nonnegative integers, and x is at most 2 or y is at least 3?
+21 There are 21 solutions to the equation u + v + w + x + y + z = 10 where u, v, w, x, y, and z are nonnegative integers, and x is at most 2 or y is at least 3.
To solve this problem, we can use the following cases:
Case 1: x=0 and y>=3. In this case, the only solution is u+v+w+z=10. There is 1 solution in this case.
Case 2: x=1 and y>=3. In this case, the possible solutions are u+v+w+z=9, u+v+w+z=8, and u+v+w+z=7. There are 3 solutions in this case.
Case 3: x=2 and y>=3. In this case, the possible solutions are u+v+w+z=8, u+v+w+z=7, and u+v+w+z=6. There are 3 solutions in this case.
Case 4: x=0 and y<3. In this case, the possible solutions are u+v+w+x+y+z=10, u+v+w+x+y+z=9, u+v+w+x+y+z=8, u+v+w+x+y+z=7, u+v+w+x+y+z=6, u+v+w+x+y+z=5, u+v+w+x+y+z=4, and u+v+w+x+y+z=3. There are 8 solutions in this case.
The total number of solutions is 1+3+3+8=21
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