## anonymous one year ago The number of triangles whose vertices are the the vertices of the vertices of an octagon but none of whose sides happen to come from the sides of the octagon.

1. anonymous

2. hartnn

|dw:1444576861098:dw| Lets take the vertices of the octagon as A,B,C,D,E,F,G,H

3. hartnn

the logic is, we can't take adjacent vertices of octagon to be vertices of the triangle. so we can't have A,B,C , but we can have A,C,E

4. anonymous

How many of such triangles would be there?

5. hartnn

thats just the first step... Pick 2 vertices. say A and C, (note you can't pick 2 adjacent vertices) try to find how many triangles that can e formed using these 2 vertices

6. hartnn

the logic makes sense?

7. hartnn

here are 2 lists: {A,C,E,G} and {B,D,F,H} we can choose 2 co-ordinates from the 1st list and the 3rd has to be from the 2nd list. this can be done in $$^4C_2 ~~ ways$$

8. hartnn

actually $$^4C_2 \times ^4C_1$$ ways. because there are 4 different vertices in 2nd list.

9. hartnn

same is true when you select 2 vertices from 2nd list and 1 from the 1st. so in all, there will be $$\Large 2 (^4C_2 \times ^4C_1)$$ ways :)