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There are n triangles of positive area that have one vertex A(0, 0) and the other two vertices whose coordinates are drawn independently with replacement from the set {0, 1, 2, 3, 4} e.g. (1, 2), (0, 1) (2, 2) etc. What is the value of n?
A. 32
B. 64
C. 128
D. 256
E. 512
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A. 32
B. 64
C. 128
D. 256
E. 512
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
13 Feb 2021, 19:32
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Bunuel
There are n triangles of positive area that have one vertex A(0, 0) and the other two vertices whose coordinates are drawn independently with replacement from the set {0, 1, 2, 3, 4} e.g. (1, 2), (0, 1) (2, 2) etc. What is the value of n?
A. 32
B. 64
C. 128
D. 256
E. 512
A. 32
B. 64
C. 128
D. 256
E. 512
We need to follow two rules, the first one is no points are allowed to be the same. The second one is we cannot have any points on the same line. Hence B(1, 1) and C(2, 2) cannot be included.
There are 25 points from 5*5 but we cannot include (0, 0). Thus we want to choose any 2 points from the rest of the 24. \(24C2 = 24*23/2 = 12*23 = 276\).
Next, we eliminate the points on the same line. Any 2 points from (x, x) cannot be chosen so that is \(4C2 = 6\) options ruled out. Don't count (0, 0) as one of the points since we already discluded it. We also have (1, 2) + (2, 4), and (2, 1) + (4, 2) so that is 2 more triangles. Finally, we cannot have the other 2 points both on the x-axis or y-axis so we ruled out \(4C2 * 2 = 12\) triangles. Therefore we take out \(12 + 6 + 2 = 20\) triangles in total.
\(276 - 20 = 256\).
Ans: D
28 Jun 2022, 10:12
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