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Selecting 3 persons sitting round a table

  • Thread starter Saitama
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  • #26
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Consider 6 people sitting round the table.
Why do you use n, if you consider n=6 only?

For now I am considering only those cases when the second person is one seat away from the first.
You mean with 1 seat between them?

I agree with verty, please use a bigger table if you want to use specific n (and give those n!).
 
  • #27
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Yes, 6*2*1=12 ways. But if the three people selected are A, E and C then the 12 counts selections of them in different orders as different. You don't really care what order. You need to divide by something.
Divide by 2?

Is this correct for n people too? I mean, for n people would there be n*2*1/2 ways?

Why do you use n, if you consider n=6 only?
I don't consider n=6 only, I used it in my previous post just to explain my thinking. Sorry if that confused you more. :redface:

mfb said:
You mean with 1 seat between them?
Yes.

mfb said:
I agree with verty, please use a bigger table if you want to use specific n (and give those n!).
For n too, its n*2*1/(something), I don't see how this is wrong. :confused:
 
  • #28
Dick
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Divide by 2?

Is this correct for n people too? I mean, for n people would there be n*2*1/2 ways?
Why divide by 2?? How many ways can you order 3 people? And no, it's not n*2*1 for any n. Why would you think that? And once you get this case, don't forget there is another one.
 
  • #29
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I think you should NOT consider n=6, because the table is too crowded. Save it for a test case after you have the general formula, and see why it works in that case. Try say n=9 instead.

To place the first person (in ordered fashion for now), there are n options of course.

To place the second person - how many options?

To place the third person - how many options? (two distinct cases based on the second person's location - you will need the probabilities of each).

Then divide out the repeated possibilities (because this is not ordered selection).
 
  • #30
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Hello Pranav,
You can skip the casework too.Just notice that after selecting all the persons required(=3) you are left with n-3 persons and they are distributed in the gaps between adjacent members (there are 3 gaps too right?) ,each gap being nonempty.So you are left with the job of calculating the number of ways in which you can distribute n-3 things in 3 groups with each being non empty .How many ways are there to do this? Now to make such a selection you need to have chosen how many members before?(Thing is that there are n different persons.) .The constant is just a consequence of counting the same thing again in this method.Can you guess what it is?
Please correct me if I am wrong.
Regards
Yukoel
 
  • #31
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I think the approach with the gaps gives casework elsewhere, as you have to fit that "gap pattern" to the table. (3,3,3) in the gaps (on a table of 12) give 4 options, (3,3,4) in the gaps (on a table of 13) give 13 options, ...
 
  • #32
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I think the approach with the gaps gives casework elsewhere, as you have to fit that "gap pattern" to the table. (3,3,3) in the gaps (on a table of 12) give 4 options, (3,3,4) in the gaps (on a table of 13) give 13 options, ...
Hello mfb,
I think we can skip the casework too .I meant that one can see the remaining n-3 members have to be distributed after selecting the 3 required.(So as to fit in the table).Selecting just one member allows for the gaps to be numbered according to the number of elements within(As you say like 3,3,3 or 3,4,4 )(I am sticking to counting the gaps anticlockwise).Then I am thinking of the usage of the formula for distribution of say k objects into r groups so that none remains empty. This method can be used for more than 3 persons to be selected as well(r changes).
Am I wrong in here?
Please do tell. Thanks in advance.
Regards
Yukoel
 
  • #33
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If you select 3, you are done. The seating order of the members is fixed. You can select 1 and treat (3,4,3) and (3,3,4) as different cases afterwards. Hmm... that does work without case-by-case analysis, but then you have to evaluate a sum over n summands.
 
  • #34
verty
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Hello mfb,
I think we can skip the casework too .I meant that one can see the remaining n-3 members have to be distributed after selecting the 3 required.(So as to fit in the table).Selecting just one member allows for the gaps to be numbered according to the number of elements within(As you say like 3,3,3 or 3,4,4 )(I am sticking to counting the gaps anticlockwise).Then I am thinking of the usage of the formula for distribution of say k objects into r groups so that none remains empty. This method can be used for more than 3 persons to be selected as well(r changes).
Am I wrong in here?
Please do tell. Thanks in advance.
Regards
Yukoel
When n = 6, we have only 111 with 4 copies. I mean, we have 6 ways to select the first seat, but 4 are copies with this pattern. When n = 7, we have 112 with 0 copies. When n = 8, 113 and 122 with 0 copies.

When 3 divides n - 3, we get a pattern with copies, 111 or 222 when n = 9. So you have to subtract these I think.
 
  • #35
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4 copies? Do you mean 3?
When 3 divides n - 3, we get a pattern with copies, 111 or 222 when n = 9. So you have to subtract these I think.
Hmm, well, it is possible.
 
  • #36
verty
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4 copies? Do you mean 3?
Only 2 should be counted, is what I meant. When n=6, we have ACE or BDF, even if C was selected.
 
  • #37
haruspex
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Guys, this problem really isn't that hard.
As I said in post #18, consider each person as taking two adjacent seats, keeping a vacant seat to the left, say. The first pair can go in n positions. That leaves a line of n-2 seats, numbered 1 to n-2 from the left, say. In how many places can the leftmost of the remaining two pairs go, remembering to leave space for the last pair? If they take seats 1 and 2, how many choices for the last pair? If they take seats r and r+1, how many choices for the last pair?
 
  • #38
Dick
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Guys, this problem really isn't that hard.
As I said in post #18, consider each person as taking two adjacent seats, keeping a vacant seat to the left, say. The first pair can go in n positions. That leaves a line of n-2 seats, numbered 1 to n-2 from the left, say. In how many places can the leftmost of the remaining two pairs go, remembering to leave space for the last pair? If they take seats 1 and 2, how many choices for the last pair? If they take seats r and r+1, how many choices for the last pair?
Pretty much what I've been saying from the beginning. The direct way actually easy if you just break into two cases depending on where the second person is seated.
 
  • #39
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If you select 3, you are done. The seating order of the members is fixed. You can select 1 and treat (3,4,3) and (3,3,4) as different cases afterwards. Hmm... that does work without case-by-case analysis, but then you have to evaluate a sum over n summands.
Well can't we just use (n-3-1)C(3-1) ?(The number of distributing n things in r groups with each of the being nonempty is (n-1)C(r-1) )Here n=n-3 and r=3.
When n = 6, we have only 111 with 4 copies. I mean, we have 6 ways to select the first seat, but 4 are copies with this pattern. When n = 7, we have 112 with 0 copies. When n = 8, 113 and 122 with 0 copies.

When 3 divides n - 3, we get a pattern with copies, 111 or 222 when n = 9. So you have to subtract these I think.
Maybe I misunderstand the problem but the people are treated as distinct right?
So the copies of the same would be for n=6
ACE;CEA;and EAC(The counting is done clockwise only and so is the choosing of gaps).Hence there are three copies of the same thing right? Well yes one can subtract the overcounted things or maybe divide it as well by 3.What I am saying is to fix the first member on the circle and proceed anticlockwise through it with gaps.I do not understand the case with n=7 'cause each single choice has 3 copies.(Something like ACE again)
For n=9 111 doesn't fit on the table so it isn't equivalent to 222 which does fit.Is this wrong ?
Please do tell.
Thanks
Yukoel
 
Last edited:
  • #40
verty
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Maybe I misunderstand the problem but the people are treated as distinct right?
Too much thinking for me, sorry. I don't like how the sum depends on n. Let's leave it.

Using cases is definitely quickest. I guess this is a general principle - try cases first.
 
  • #41
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I was confused by so many replies so I did not reply for sometime but I did figure it out. It will be difficult for me to put it in words but I will try my best.

Selecting three persons without restriction, nC3.
When three persons are sitting together, there are n ways.
In the last case, when two persons are sitting together, there can be n ways to select two persons sitting next to each other and the last person could sit at any of (n-4) places as he cannot sit near any of the two persons.

Hence, total ways=nC3-n-n*(n-4). Solving, I do end up with a right answer.

Thank you everyone! :smile:
 

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