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?\[\]

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TopNewestOkay 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 · 1 year, 3 months ago

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– Surya Prakash · 1 year, 3 months ago

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

– Satyajit Ghosh · 1 year, 3 months ago

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

– Surya Prakash · 1 year, 3 months ago

yesLog in to reply

– Satyajit Ghosh · 1 year, 3 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.?Log in to reply

– Surya Prakash · 1 year, 3 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.Log in to reply

– Satyajit Ghosh · 1 year, 3 months ago

Thanks \(\ddot\smile\)Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

Is mine correct?Log in to reply

– Satyajit Ghosh · 1 year, 3 months ago

Is your and my answer the same?Log in to reply

@Adarsh Kumar @Surya Prakash is it flawed? – Satyajit Ghosh · 1 year, 3 months ago

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@Souryajit Roy – Adarsh Kumar · 1 year, 3 months ago

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The same question came in my Jharkhand region with a slight variation - 32 instead of 28. – Ankit Kumar Jain · 1 year, 3 months ago

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– Satyajit Ghosh · 1 year, 3 months ago

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

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. 2

21+222+20*20=486 243 due to repetitions Multiply bi 28 ÷ by 3 to get2268 – Ajinkya Shivashankar · 1 year, 3 months agoLog in to reply

– Ajinkya Shivashankar · 1 year, 3 months ago

Forgot to divide by 28(as it is circular)Log in to reply

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 · 1 year, 3 months ago

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– Adarsh Kumar · 1 year, 3 months ago

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

– Surya Prakash · 1 year, 3 months ago

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

– Adarsh Kumar · 1 year, 3 months ago

Surya,is it correct?Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

Oh,no problem buddy!Log in to reply