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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?

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  • $\begingroup$ Your account has the same name as this other one- did you accidentally make a new one? $\endgroup$ – Kevin Long May 2 '18 at 21:03
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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.

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From any given vertex you can draw $7$ diagonals. You can't draw to itself or to the two neighboring points. There are therefore $10 \cdot 7$ ends of diagonals. As each diagonal has two ends, there are $10 \cdot 7 \cdot \frac 12=35$ diagonals.

The approach to the formula you quote is that you pick two vertices to draw a line between, which you can do in $10 \choose 2$ ways. $10$ of those are sides of the decagon instead of diagonals, so the result is ${10 \choose 2}-10=35$

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The correct answer gives you $\binom{10}{2}-10=35$, the reasoning being that you choose two vertices of the $10$ and draw a diagonal between them. However, as you said, consecutive vertices will not give a diagonal, so we ignore the $10$ sides of the decagon that arise from this counting.

I have trouble telling what you're doing in your answer; you seem to fix a vertex, but there shouldn't be $8$ possible vertices you can draw a diagonal to from that fixed vertex, but $7$, because you subtract both vertices adjacent to the fixed one, but also the fixed vertex itself. There's also no reason to use a binomial coefficient there. So, for each vertex, you can pick $7$ possible vertices to draw a diagonal to that one, and since there are $10$ vertices, you should get $7*10$. However, this counts each diagonal twice- once for each vertex on that diagonal, so you divide by $2$. This gives you $35$, the same answer as before.

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All answers are correct, I thought I can throw another analogy to your question. Assume 10 people get into a room and they shake their hands. How many handshakes are there? Answer: 10c2 = 45; OR (better intuitive approach) if I'm in the room, I will shake hands to 9 people. Since we don't count the same handshake twice we'll divide it by 2. So we can say: $10 \cdot 9 \cdot \frac 12=45$

Back to your question: if 10 people get into a room and they are seated in a circle but they don't shake the hand of the person on the left and on the right (and not to themselves of course), how many handshakes we'll have?

Answer: $10 \cdot 7 \cdot \frac 12=35$

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Choose a vertex ($10$ possible ways) join it to any of the other $7$ possible vertices, this will double count the diagonals so there are $10 \times 7 /2$ possible diagonals.

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