Try the following problem. And, remember, don’t give up. If you think a certain approach might work, try it. Giving up, however, will never get you the answer.
A twelve-sided polygon consists of vertices A – L. How many lines can be drawn between any two vertices, such that a line is neither repeated, nor redundant, with any sides of the polygon?
Find a Pattern
Start small. That’s right. Instead of a massive dodecahedron (the name of a twelve-sided polygon) ask yourself the same question, but with a square. A square will yield two such lines. And, a pentagon, a five-sided figure, will yield five such lines (you can quickly draw this out on a scratch paper). Finally, a hexagon will yield nine such lines.
At this point, you should notice that we are adding a number that was one greater than the number we added in the previous case. For the pentagon we added 3 to the total, for the hexagon, 4. For a seven-sided figure, we would therefore add another 5. For an octagon, 6.
Now, we have a sequence of numbers 2 + 3 + 4 + 5 + 6…. notice that the final number I am adding is always two less than the number of sides. Therefore, for a twelve-sided figure, I will have to continue the series all the way up to 10: 2 + 3 + 4… + 9 + 10 = 54. And there’s our answer: (B).
A Totally Different Method
This is a great, intuitive way to solve problems. Of course, it is not the most elegant way, but requires some brute force. But that’s okay – as long as you get the answer in a reasonable amount of time, that is the important part.
But, let’s think of this problem in a totally different way. I am choosing from twelve letters. Each line has two vertices; so, therefore, I am choosing two from 12. The order doesn’t matter – meaning that line AG is the same as line GA.
Now, we use the combinations approach. And, notice I say approach, and not formula, because the formula for combinations is needlessly cumbersome and potentially intimidating (watch my videos on Combinations/Permutations).
Using the combinations approach we get: (12)(11)/2 = 66. Now, we have to account for the redundancies. Meaning that there are lines which are already part of the polygon, i.e. we wouldn’t be drawing a line from A to B, because Line AB is already part of the polygon. How many lines are there in a twelve-sided figure? 12, of course.
So, 66 -12 = 54. The answer.
Now, we can turn this into a formula, but I am not going to do so. The key to the GRE is not to cram a bunch of formulas into your head, most of which you’ll probably never see. The key is how to approach problems. In this case, we found a pattern, and, from that pattern, we were able to extrapolate an answer. Developing that skill is far more important than memorizing formulas.