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How many unordered pairs of edges of a given cube determine a plane?
a. 12. b. 28 c. 36 d. 42. e. 66
I ended up with the right answer. However, no solutions resembled the way I did it, which is why I'm skeptical. After a few failed attempts with other ways of solving (didn't get an answer in the answer options), I decided to use complementary counting, and counted invalid pairs (pairs that don't satisfy the condition given by the problem). To do that I drew out the cube and found for each side, there are 4 sides that will make an invalid pair with it. Because there are 12 sides to a cube, I did 12*4. But, each distinct pair in that 12*4 is counted twice. So, dividing 12*4 by 2 will give 24 invalid pairs. Now, for total pairs, it is 12 choose 2, which is 66. 66-24 = 42. Hence, the answer is 42.
So, is there anything wrong with the way I did it? Is there a specific reason this wasn't a posted solution?
Thanks!
a. 12. b. 28 c. 36 d. 42. e. 66
I ended up with the right answer. However, no solutions resembled the way I did it, which is why I'm skeptical. After a few failed attempts with other ways of solving (didn't get an answer in the answer options), I decided to use complementary counting, and counted invalid pairs (pairs that don't satisfy the condition given by the problem). To do that I drew out the cube and found for each side, there are 4 sides that will make an invalid pair with it. Because there are 12 sides to a cube, I did 12*4. But, each distinct pair in that 12*4 is counted twice. So, dividing 12*4 by 2 will give 24 invalid pairs. Now, for total pairs, it is 12 choose 2, which is 66. 66-24 = 42. Hence, the answer is 42.
So, is there anything wrong with the way I did it? Is there a specific reason this wasn't a posted solution?
Thanks!