anonymous
  • anonymous
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.
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
@hartnn Can you please help?
hartnn
  • hartnn
|dw:1444576861098:dw| Lets take the vertices of the octagon as A,B,C,D,E,F,G,H
hartnn
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
How many of such triangles would be there?
hartnn
  • 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
hartnn
  • hartnn
the logic makes sense?
hartnn
  • 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 \(^4C_2 ~~ ways\)
hartnn
  • hartnn
actually \(^4C_2 \times ^4C_1\) ways. because there are 4 different vertices in 2nd list.
hartnn
  • 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 :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.