Combinatorics-RMO 2015

Suppose there are 36 equally spaced points on the circumference of a circle.In how many ways you can choose 3 any two points such that no 2 of them are adjacent or diametrically opposite?

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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We shall use PIE.

Step 1: Number of ways of selecting $3$ points from $36$ points is $\binom{36}{3} = 7140$ .

Step 2: Number of ways of selecting $3$ adjacent points is $36$ .

Step 3: Number of ways of selecting $2$ adjacent and one not adjacent with them is $36\times 32 = 1152$ . (Since there are $32$ ways to select the non-adjacent point.)

Step 4: Number of ways of selecting two diametrically opposite points are $18$ and number of ways of selecting third one not adjacent to both of them are $30$ in each case. So total number of ways in this step are $18 \times 30 = 540$ .

Step 5: Number of ways (what we required) = Total $-$ Number of ways of selecting $3$ adjacent points $-$ Number of ways of selecting $2$ adjacent and one not adjacent with them - Number of ways of selecting two diametrically opposite points and selecting third one not adjacent to both of them $= 7140-36-1152-540=5412$ .

Surya Prakash - 4 years, 10 months ago

I think there were 32 points instead of 36.

Prakher Gaushal - 4 years, 10 months ago

I have given RMO of West Bengal Rgion and there was 36 points in the problem (problem no 4)....Perhaps in your region they have given 32....basically the procedure of solving the sum is not affected if the number of points in even.

Souryajit Roy - 4 years, 10 months ago

5430

Prakher Gaushal - 4 years, 10 months ago

For 32 objects you can have a look at my solution here

Shubhendra Singh - 4 years, 10 months ago

What can be expected cutoff for rmo

satyam mani - 4 years, 10 months ago

around 55

Raushan Sharma - 4 years, 10 months ago

bro 28

Priyansh Gupta - 2 years, 9 months ago

The answer is 5400.

Let the points be A,B and C. We can put A anywhere on the circle, giving 36 ways.

Then, B and C cannot be on the two points next to A, or the point opposite it, leaving 32C2 = 496 ways to choose B and C.

Of these, 30 configurations have B and C adjacent, and 16 have them opposite. We subtract these off to give 450 valid ways.

For each valid configuration, there is one way to generate it starting with A at each vertex. Thus each configuration has been counted 3 times, so we divide by 3 to get the final answer:

36x450/3 = 5400

Daniel Tan - 4 years, 8 months ago

What was your answer?

Prakher Gaushal - 4 years, 10 months ago

32*36

Dipanjan Chowdhury - 4 years, 10 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}`	$\boxed{123}$