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# RMO 2015 Telangana Region Combinatorics question.

Suppose 28 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?Here is my solution,could somebody please tell if it is correct not?

Note by Adarsh Kumar
1 year, 10 months ago

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Okay this is my method which I guess is hopefully correct. This is just an outline. Consider a polygon of 28 sides. So a diagonal is a line such tha it touches no 2 adjacent vertex. Now if we subtract n/2 from the no. Of diagonals i.e. n(n-3)/2 we get the no. Of such options available. (If you are not sure you can check out this thing that n(n-3) -n/2 always gives diagonals subtracting diametrically opposite objects with 4,6,8,10 sided polygon. Now from this we get 336c3.

- 1 year, 10 months ago

But you should prove that number of diagonals in an '$$n$$' sided polygon are $$n(n-3)/2$$.

- 1 year, 10 months ago

I wrote this answer on my phone so had to shorten it. Btw is it correct?

- 1 year, 10 months ago

yes

- 1 year, 10 months ago

thanks. But do we have to prove that no. Of diagonals is n(n-3)/2 in the rmo paper? And do we have to write like step 1,step2 etc.?

- 1 year, 10 months ago

Yes, You should write the proof that no. of diagonals is $$n(n-3)/2$$. And there is no need of writing step 1 and step 2. Just the point is that you should explain your solution clearly, I mean you should elaborate your solution.

- 1 year, 10 months ago

Thanks $$\ddot\smile$$

- 1 year, 10 months ago

Is mine correct?

- 1 year, 10 months ago

Is your and my answer the same?

- 1 year, 10 months ago

@Adarsh Kumar @Surya Prakash is it flawed?

- 1 year, 10 months ago

The same question came in my Jharkhand region with a slight variation - 32 instead of 28.

- 1 year, 10 months ago

The same question came in mostly all the states with a little variation like delhi had 36.

- 1 year, 10 months ago

Use brute force If you chose 1 & 3(or 27) you can chose 21 others If you chose 1 & 13(or 15) you can chose 22 others If you chose 1 and anything else you can choose 20 others. 221+222+20*20=486 243 due to repetitions Multiply bi 28 &divide by 3 to get2268

- 1 year, 10 months ago

Forgot to divide by 28(as it is circular)

- 1 year, 10 months ago

Step 3 is wrong @Adarsh Kumar , Because we should choose two diametrically opposite points and the third point not adjacent to it. As the case when third point is adjacent is already included in step two.

- 1 year, 10 months ago

That is why I have added the ways common to step 2 and 3 in step 4.

- 1 year, 10 months ago

Ohh Sorry!! I didn't observe that. I just saw till the middle and left that as I thought there is mistake.

- 1 year, 10 months ago

Surya,is it correct?

- 1 year, 10 months ago

Oh,no problem buddy!

- 1 year, 10 months ago