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10 persons are sitting in a circle. In how many ways can you select any three of them such that no two of them are consecutive?

My attempt: To get the required number, we subtract the unfavourable cases from the total number of cases.

The total number of cases will be the number of ways to select three people out of 10 people - 10C3 = 120

Out of these cases, unfavourable cases are those in which

(1) two of our selected people sit together while the third is sitting apart.

First we select any one of those people in 10C1 ways. To select a person sitting next to this person, we have 2C1 ways. Now, there are only 6 ways to select the third person. Therefore, the total number of ways is $10*2*6=120$.

(2) All three of them are sitting together:

There are 10 ways to do that.

The total number of unfavourable cases = 10+ 120 = 130

My unfavourable cases are exceeding the total number of cases. What am I missing?

Parcly Taxel
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Arishta
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2 Answers2

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$\underline{A\; direct\; method}$

I take it that it is a numbered circle.

Make $3$ blocks of chosen-unchosen, looking clockwise, $\boxed{\circ\bullet}$ so there are now $3$ blocks and $4$ "singles"

The blocks can be placed among the $7$ units in $\binom73 = 35$ ways,

but each unit is getting only $7$ starting points instead of $10$,

thus the final answer $= \dfrac{10}{7} \times 35 = 50$

true blue anil
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  • Hi. What is meant by "but each unit is getting only 7 starting points instead of 10"? Any link for understanding this part better? Thanks. – aarbee Sep 03 '20 at 17:11
  • @aarbee: Sorry, I was on tour while posting the answer, and saw your query only now. The circle has been compressed from $10$ points to $7$ points due to forming blocks, so the "chosen" now have fewer "numbered" points to start from. See the formula for dominoes in a circle in the classic paper https://math.dartmouth.edu/~doyle/docs/menage/menage/menage.html – true blue anil Aug 19 '23 at 14:24
  • Thank you :-).. – aarbee Aug 19 '23 at 16:31
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Selecting the people one at a time is, in general, never the correct way. To deal with case (1), the following should be done instead:

  • Select the block of two: 10 ways
  • Select the last person so that a block of three isn't formed: 6 ways

So that gives 60 ways. The handling of case (2) is correct, so there are $120-60-10=50$ valid selections.

Parcly Taxel
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  • There are 10 ways to select the first person. There are two ways to make the next selection and 6 ways to make the third selection. Can you tell me which cases did I overcount when I selected the people one at a time? – Arishta May 07 '17 at 17:31
  • @Cotton Suppose the first person selected is labelled A and the second one B. Selecting B first followed by A is the same thing, but you treated them as different. You would have to divide 120 by 2 there. – Parcly Taxel May 07 '17 at 17:32
  • Alright, so while selecting the people one at a time, I also considered the order in which they were selected. I get it. Thank you so much. – Arishta May 07 '17 at 17:37
  • Is the seating order considered different. For example – Peter May 07 '17 at 18:00
  • @Peter No ${}{}$ – Parcly Taxel May 07 '17 at 18:01
  • So if person A is sitting to the left OR the right of person B, this is considered the same – Peter May 07 '17 at 18:04
  • @Peter Yes. I consider them as a block of two to avoid having to divide. – Parcly Taxel May 07 '17 at 18:04
  • What about in a clockwise direction ABCDE and DCABE. Are they considered the same – Peter May 07 '17 at 18:06
  • @Peter If they are the same group of people then yes. – Parcly Taxel May 07 '17 at 18:08