What is the probability that the selected triangle has no side common with the octagon?

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Solution 3 (Stars and Bars)
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From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

Solutions

Solution 1

We will use constructive counting to solve this. There are cases: Either all points are adjacent, or exactly points are adjacent.

If all points are adjacent, then we have choices. If we have exactly adjacent points, then we will have places to put the adjacent points and also places to put the remaining point, so we have

choices. The total amount of choices is

Thus our answer is

Solution 2

We can decide adjacent points with choices. The remaining point will have choices. However, we have counted the case with adjacent points twice, so we need to subtract this case once. The case with the adjacent points has arrangements, so our answer is

Solution 3 (Stars and Bars)

Let point of the triangle be fixed at the top. Then, there are

ways to chose the other 2 points. There must be spaces in the points and points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and extra points (k-1) distributed so by the stars and bars formula,

, there are

ways to arrange the bars and stars. Thus, the probability is