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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.

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  1. anonymous
    • one year ago
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    @hartnn Can you please help?

  2. hartnn
    • one year ago
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    |dw:1444576861098:dw| Lets take the vertices of the octagon as A,B,C,D,E,F,G,H

  3. hartnn
    • one year ago
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    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
    • one year ago
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    How many of such triangles would be there?

  5. hartnn
    • one year ago
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    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
    • one year ago
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    the logic makes sense?

  7. hartnn
    • one year ago
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    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
    • one year ago
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    actually \(^4C_2 \times ^4C_1\) ways. because there are 4 different vertices in 2nd list.

  9. hartnn
    • one year ago
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    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 :)

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