I have a statement that says:
How many diagonals can be traced in a regular decagon (10-sided
polygon)?
My development was:
First the diagonals, are drawn to non-consecutive vertex, so are $8$ available vertex. And we want to select 2 vertex, the fixed point and the point to which I will draw the diagonal.
So i use: $\frac{n!}{(n – r)! * r}$, also for each vertex, I do not take into account the fixed vertex, this is done 10 times, so I will subtract 10.
Then, the my final result is $\frac{8!}{(8-2)! * 2!} – 10$,
but the correct result must be $\frac{10!}{(10 – 2)! * 2!} – 10$, Where is my error?
Best Answer
It does not follow that there are then $\frac{8!}{(8-2)!\cdot 2!}$ ways. It is essentially just a non-sequitur. When you subtract $10$, it doesn't mean anything.
Let me provide an argument as similar as possible to your argument.
Pick a vertex. There are $7$ other vertices we can draw to. Therefore, there are $7$ diagonals for each vertex. There are $10$ vertices total, so the total counted is $7 \cdot 10$. But for each diagonal, there are $2$ vertices that we picked. Therefore, we counted each diagonal twice! The answer is therefore
$$\frac{7 \cdot 10}{2} = 35$$
Now, let's look at what the solution intended. There are $10$ vertices. A segment is defined by its two endpoints. Therefore, there are
$$\binom{10}{2} = 45$$
segments, and $10$ of these are the sides of the decagon. Therefore there are $45 - 10 =35$ diagonals.