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

#2**+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 v_{i} to v_{j} is the same diagonal drawn from v_{j} to v_{i}, 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**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

#5**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