Melody

ok I have a different idea.

There are 7C2 = 21 possible cords in total

I want 4 of them so there are 15C4 = 1365 ways to select 4 chords

To make a bow tie I need to chose just 4 points from the 7 so that is 7C4=35 ways

With each of these 35 sets of 4 points there is 2 ways to join them to make a bow tie, so that is 70 bow ties.

So maybe the answer is    70/1365 = 2/39

Can you please check that answer.

I am assuming that you have the ability to plug numbers into a box and be told right or wrong.  Can you do that?

(rotations are not counted as the same)

So what is the answer?

trianlges  10C3 = 120

quadrialterals 10C4 = 210

pentagons  10C5 = 252

hexagons 10C6 = 210

heptagons 10C7 = 120

octogons 10C8 = 45

nonogons 10C9 = 10

deacagons 10C10 = 1

120+210+252+210+120+45+10+1 = 968 cyclic polygons

But then the question only wants the quadrilaterals so I get  210

All the terms from 10! have  2*5*10 = 100 as a factor so none of these will affect the last 2 places in the sum

So I only need the last 2 places of 5!

The last place will be 0  (2*5)

1*2*3*4*5 = 120 

So the last 2 digits are 20

So my answer is confirmed as correct?

continued here

Please include the link to your first post.

Spongebob, 

It is really rude of you to give yourself 6 points and geno no points at all.  You gave him no thanks either.

Do you think this rude behaviour helps you in some way?  Do you think no one will notice?

\(\frac{cos(-\theta)cot(-\theta)}{cosec(\theta)}\\ =cos(\theta)*\frac{cos(-\theta)}{sin(-\theta)}\div\frac{1}{sin(\theta)}\\ =cos(\theta)*\frac{cos(\theta)}{-sin(\theta)}*\frac{sin(\theta)}{1}\\ =-cos^2(\theta)\)