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?

  • $\begingroup$ what about when three people are sitting together? $\endgroup$ – Angina Seng May 7 '17 at 17:15
  • $\begingroup$ Oh yeah, I forgot that. I will do the calculation again and make the edits. $\endgroup$ – Arishta May 7 '17 at 17:17
  • $\begingroup$ @LordSharktheUnknown I have made the edits. Would you take a look at it now? $\endgroup$ – Arishta May 7 '17 at 17:20
  • 1
    $\begingroup$ Possible duplicate of Combinatorics: Selecting objects arranged in a circle $\endgroup$ – user940 May 7 '17 at 17:22
  • $\begingroup$ Your argument (1) can't be correct: are you sure you haven't counted some possibilities more than once? $\endgroup$ – Angina Seng May 7 '17 at 17:22

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.

  • $\begingroup$ 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? $\endgroup$ – Arishta May 7 '17 at 17:31
  • $\begingroup$ @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. $\endgroup$ – Parcly Taxel May 7 '17 at 17:32
  • $\begingroup$ 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. $\endgroup$ – Arishta May 7 '17 at 17:37
  • $\begingroup$ Is the seating order considered different. For example $\endgroup$ – Peter May 7 '17 at 18:00
  • $\begingroup$ @Peter No ${}{}$ $\endgroup$ – Parcly Taxel May 7 '17 at 18:01

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

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 $= \dfrac{10}{7} \times 35 = 50$

  • $\begingroup$ 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. $\endgroup$ – aarbee Sep 3 '20 at 17:11

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