# How many pentagons are there such that all their vertices are points from this set of twelve points and each of the pentagons has at least on vertex of each colour?This question is a tricky...And...

How many pentagons are there such that all their vertices are points from this set of twelve points and each of the pentagons has at least on vertex of each colour?

This question is a tricky...And thank you so much for trying to help!

Twelve points are placed on the circumference of a circle. Five of these twelve points are coloured blue and the other seven are coloured red. HOW MANY PENTAGONS ARE THERE SUCH THAT ALL THEIR VERTICES ARE POINTS FROM THIS SET OF TWELVE POINTS AND EACH OF THE PENTAGONS HAS AT LEAST ONE VERTEX OF EACH COLOUR?

I am really confused so would you please leave all working! Thank you...

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### 1 Answer

(1) Assume that by pentagon you mean a 5-sided convex polygon.

(2) Assume that any 5 points determines exactly 1 pentagon -- e.g. ABCDE and ACDBE name the same pentagon. In other words, the pentagon is described by the vertices in any order.

There are `_12C_5=792` different pentagons that can be formed. (`_nC_r` is the number of combinations of n things taken r at a time without replacement) You can also see this by using the fundamental counting principal. You have 12 choices for the first vertex, 11 for the second, etc... so you have `12*11*10*9*8=95040` pentagons. But since the order that the vertices is listed doesn't matter, for each pentagon you must divide by the number of permutations which is `5*4*3*2*1` ; thus the total number of pentagons is `95040-: 120=792`

There are `_7C_5=21` different pentagons that are all red. There are `_5C_5=1` pentagon that is all blue. All other pentagons have at least 1 of each color.

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**The number of pentagons that can be formed with at least one vertex of each color is 792-22=770.**

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