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#1 2006-02-21 22:40:34
- rimi
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- Registered: 2006-02-21
- Posts: 17
Same old permutation-combination
Here goes a simple permutation and combination puzzle..
There is a polygon with 10 sides.A point is there in the middle of the polygon and all the vertices are joined with the point and therefore producing 10 triangles.In exactly how many ways can you select three triangles no two of which are adjacent?
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#2 2006-02-21 23:41:51
- Jai Ganesh
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- Registered: 2005-06-28
- Posts: 48,422
Re: Same old permutation-combination
10 x 7 x 4 = 280 
Thats a guess 
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#3 2006-02-22 09:37:45
- mathsyperson
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- Registered: 2005-06-22
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Re: Same old permutation-combination
Brxxt! Wrong!
If the first two triangles you select are as close to each other as you can get them, you get more choice for the third one, meaning that you can choose 300 in all.
But because of rotational and reflectional symmetry, only 15 of these are unique.
Why did the vector cross the road?
It wanted to be normal.
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#4 2006-02-22 16:55:12
- rimi
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- Registered: 2006-02-21
- Posts: 17
Re: Same old permutation-combination
I repeat my question ....there is a point within a polygon of 10 sides , suppose ABCDEFGHIJ...take a point inside , say O.and join the vertices with this point producing 10 triangles,viz, AOB,BOC,COD,DOE,EOF,FOG,GOH,HOI,IOJ,JOA.
now you have got to select three triangles out of these 10 so that no two are adjacent.
Ganesh you are not right...there are actually less choices.
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