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# How many diagonals can be drawn in the pentagon?

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How many diagonals can be drawn in the pentagon?
(regular pentagon, nothing special)
A: 5

B: 10

C: 15

D: 20

(This chapter is about Permutations and Combinations)

Guest May 8, 2014

### Best Answer

#1
+5453
+8

There are 5 diagonals, no repeats. For example, AD and DA are the same diagonal, so it should be a combination.

kitty<3  May 8, 2014
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### 7+0 Answers

#1
+5453
+8
Best Answer

There are 5 diagonals, no repeats. For example, AD and DA are the same diagonal, so it should be a combination.

kitty<3  May 8, 2014
#2
+84340
+5

I think the number of diagonals that can be drawn in any poygon with n vertexes (sides), where n >3, is given by: (n)(n-3)/2.

To see this, note that no diagonal can be drawn from a vertex to itself, nor can any diagonal be drawn to the two neighboring vetexes. So this leaves (n-3) vertexes left. And from each vertex, the same number of diagonals can be drawn. So the total number of diagonals that can be drawn is (n)(n-3). But the diagonal from vi to vj is the same diagonal drawn from vj to vi, so we must divide this total number by two to prevent "double counting."

So we have...... (n)(n-3) / 2.

CPhill  May 8, 2014
#3
+91972
0

Thank you Kitty, that is a good answer.

However, I would also like to see this done using combination notation.

Melody  May 8, 2014
#4
+91972
0

Thank you Chris, you must have been reading my mind.

Melody  May 8, 2014
#5
+84340
0

P. S.........there may be some combinatoric that could be found for this, but I couldn't think of any other way to do it!! Anyone else have any ideas??

CPhill  May 8, 2014
#6
+5

n=3: the number of diagonals = 0 + 1 + 2   -  3

n=4: the number of diagonals = 0 + 1 + 2 + 3  -  4

n=5: the number of diagonals = 0 + 1 + 2 + 3 + 4  -  5

n=6: the number of diagonals = 0 + 1 + 2 + 3 + 4 + 5 -  6

...

n: the number of diagonals = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + (n-1)  -  n

the number of diagonals = $$\frac{(1+(n-1))*(\;((n-1)-1)\;+\;1\;)}{2} - n = \frac{n(n-1)}{2} - n=\binom{n}{2}-n$$

the number of diagonals = $$\boxed{\binom{n}{2}-n=\frac{n(n-3)}{2}}\quad n\ge3$$

S.

Guest May 8, 2014
#7
+91972
+3

Thank you anonymous.  This has been a really interesting question.

Thank you to Chris and Kitty as well!

Melody  May 8, 2014

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