combination and permutation problem

bhupinder kaur gill · 2009-05-30 15:38:23

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.

A) 112
B) 120
C) 468
D) 710
E) None of these

please tell how to figure out this??????????????
please please

mathsyperson · 2009-05-31 03:11:41

I get 50 different ways.

The first triangle can go wherever it likes, so there are 10 choices there.
Now that triangle and its neighbours can't be chosen, so there's a choice of 7 left.

It's a bit more complicated than that though. If the second triangle is exactly 2 away from the first (which can happen in 2 ways), then there are 5 places left that the third one can go.

If the second triangle isn't 2 away from the first (which can happen in 5 ways), then there are 4 places for the last one.

Therefore, the total number of choices is 10 x (2x5 + 5x4) = 10 x 30 = 300.

However, that method counts each combination 6 times, so the actual answer is 50.

noobard · 2009-06-07 22:35:15

we can eliminate the options ...
The maximum choices will be 10c3...which is nothin but 120
so C and D are out of the picture...........It cannot be 120 also because we have our restrictions...
So we a re left with 112 and none of these............ll try figuring out the xact answer

noobard · 2009-06-08 21:13:05

i also got 50...
but ma method is a bit different

as i said earlier we got 120 triangles ....now we subtract the cases we we hav 2 adjacent and three adjacent and hence we get the answer...so for two adjacent ....we choose two adjacent triangles ...which can be in 10 ways and the third one has to be chosen from the remaining 6(10-(two chosen and two neighbours of them)..so 10*6 hence 60
and no of groups having all the three bunched is 10

so 120-60-10
which is nothin but 50

John E. Franklin · 2009-06-12 02:36:59

I drew a picture of 5 ways that are different, and then they can be
rotated 10 times, so I also get 50 like mathsy and noobard.
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
something like that.

Math Is Fun Forum

#1 2009-05-30 15:38:23

combination and permutation problem

#2 2009-05-31 03:11:41

Re: combination and permutation problem

#3 2009-06-07 22:35:15

Re: combination and permutation problem

#4 2009-06-08 21:13:05

Re: combination and permutation problem

#5 2009-06-12 02:36:59

Re: combination and permutation problem

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