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?

#Combinatorics #RMO

Easy Math Editor

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Comments

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.

Satyajit Ghosh - 5 years, 6 months ago

@Adarsh Kumar @Surya Prakash is it flawed?

Satyajit Ghosh - 5 years, 6 months ago

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

Surya Prakash - 5 years, 6 months ago

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

Satyajit Ghosh - 5 years, 6 months ago

@Satyajit Ghosh – yes

Surya Prakash - 5 years, 6 months ago

@Surya Prakash – Is mine correct?

Adarsh Kumar - 5 years, 6 months ago

@Adarsh Kumar – Is your and my answer the same?

Satyajit Ghosh - 5 years, 6 months ago

@Surya Prakash – 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.?

Satyajit Ghosh - 5 years, 6 months ago

@Satyajit Ghosh – 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.

Surya Prakash - 5 years, 6 months ago

@Surya Prakash – Thanks $\ddot\smile$

Satyajit Ghosh - 5 years, 6 months ago

@Souryajit Roy

Adarsh Kumar - 5 years, 6 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.

Surya Prakash - 5 years, 6 months ago

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

Adarsh Kumar - 5 years, 6 months ago

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

Surya Prakash - 5 years, 6 months ago

@Surya Prakash – Oh,no problem buddy!

Adarsh Kumar - 5 years, 6 months ago

@Surya Prakash – Surya,is it correct?

Adarsh Kumar - 5 years, 6 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

Ajinkya Shivashankar - 5 years, 6 months ago

Forgot to divide by 28(as it is circular)

Ajinkya Shivashankar - 5 years, 6 months ago

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

Ankit Kumar Jain - 5 years, 6 months ago

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

Satyajit Ghosh - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$