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what is the probability of 5 people with different ages sitting in ascending or descending order at a round table.

So, let me know if there's a better way to go about this problem.

Let's have the people be named 1,2,3,4,5 They could sit: 12345 23451 34512 45123 51234 or the reverse since order matters (12345 is different from 54321) There are 5! or 120 different ways the people can sit So 10/120 or 1/12 is the chance that they sit in ascending or descending order. Is there a more formal way to do this?

Also, how many ways can 5 people sit at a round table? (combination problem, order doesn't matter)

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  • $\begingroup$ Don't forget that a rounded table does not have a first sit neither a last one. $\endgroup$ – Sigur May 10 '13 at 23:13
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The answer to your second question will help illuminate a different way of thinking about the first.

If the $5$ people were to sit in a straight line, then the number of ways they can sit is indeed $5!$. Since we are asking them to sit in a circle, each possible circular arrangement corresponds to $5$ different linear arrangements, depending on which seat we label the 'first' seat. This shows that there are $5!/5=4!=24$ ways for $5$ people to sit around a table.

Now, of all $24$ possible seating arrangments, only $2$ of them are arranged in increasing or decreasing order of age, so the desired probability is $1/12$.

Notice our answers are the same, but it is important to see the distinction in our reasoning. In a sense, you've chosen a first seat, which increased your numerator and denominator by a factor of $5$. Since the question asks for a ratio, this didn't affect the final answer, but we should still note the difference between sitting around a table, and sitting in a line.

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  • $\begingroup$ I would not think of the reasoning as embodying a mistake. The chairs are all different, the view of the room is different. It is only by convention that we identify circular permutation with seating at a round table. And even if we are at the centre of a round room with all views identical, nothing prevents one from putting labels on the chairs to solve a probability problem. $\endgroup$ – André Nicolas May 10 '13 at 23:27
  • $\begingroup$ @AndréNicolas: You're absolutely right. I'm just trying to stress the convention you refer to for the sake of Jwan622's understanding of the nuances if we consider chairs to be identical. Perhaps mistake is too harsh of a word. $\endgroup$ – Jared May 10 '13 at 23:30
  • $\begingroup$ Why is it generally accepted that the chairs are identical in a round table but not in a line formation? In a round table... 12345 is different from 23451 even if the poeple are sitting in the same order... they are different formations nevertheless right? I still kind of don't see why a round table is different from a line. I guess I don't understand the nuances of this sentence: "Since we are asking them to sit in a circle, each possible circular arrangement corresponds to 5 different linear arrangements, depending on which seat we label the 'first' seat." $\endgroup$ – Jwan622 May 10 '13 at 23:37
  • $\begingroup$ @Jwan622: The answer to your question is simply a matter of convention. Usually in these problems concerning round tables, we consider two arrangements the same if they only differ by a rotation of the table. Perhaps this is because relative to the other guests, everyone is in the same position no matter how the table is rotated. This is not true of cyclic permutations for a line of chairs (ie, someone who was to the left of you could jump to the right of you if we cycle a linear arrangement). $\endgroup$ – Jared May 10 '13 at 23:44
  • $\begingroup$ I see, I guess in my problem... and the way I framed it in my head... the roundness of the table is irrelevant. I'll try to be a little more aware next time. If combination matters... can you clarify how you get 24 combinations? Logically I realize what is done... but isn't this a combination expression of this problem 5C5 = 5P5/5! = 5!/5! ? $\endgroup$ – Jwan622 May 11 '13 at 0:31
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Lets mark each seat on the round table as A, B, C, D and E.

A B C D E

1 2 3 4 5
5 1 2 3 4
4 5 1 2 3
3 4 5 1 2
2 3 4 5 1

So there are 5 ways you can sit in an ascending order and 5 ways in descending. So total of 10 ways.

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  • $\begingroup$ Normally rotations are considered to be the same since it is a circular table. $\endgroup$ – Jack Moody Aug 24 '18 at 20:08
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Your reasoning is good, and the answer is correct. I would prefer to think as follows. Call them $1,2,3,4,5$ from youngest to oldest. Sit the oldest down first. Wherever she sits, there are precisely $2$ ways the others can sit so that the $5$ people are in age order. But there are $4!$ equally likely ways for the $4$ remaining people to sit. So our probability is $\frac{2}{4!}$.

Remark: The answer to your question about the number of ways $5$ people can sit at a round table depends on context. If the question is asked in a combinatorial setting, the only answer that will be marked as correct is something equivalent to $24$. For it is the standard convention that two arrangements that differ by a rotation are considered to be identical.

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  • $\begingroup$ How did you get 24? What's odd to me is that it sounds like a combination problem of 5C5 but I thought the formula for that would be (n!/(n-k)!) / k! = 5P5/5! = 5!/ (5-5)! / 5! = 5!/1 / 5! = 5!/5! $\endgroup$ – Jwan622 May 11 '13 at 0:29
  • $\begingroup$ It is the number $4!$ of permutations of $4$ objects. $\endgroup$ – André Nicolas May 11 '13 at 0:46
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Use conditional probability. The youngest person can sit anywhere at the table, so his position doesn't matter. The second person must sit next to the first, so Let A be the event that the people at the table are in ascending order; L be the probability that the second youngest is seated to the left of the first person and R be the probability that he's on the Right. If the second youngest is not seated next to the first, all bets are off, so we'll use this as an event to condition on in order to make the problem "smaller." So p(A) = p(A,L) + p(A,R) = p(A|L)p(L) + p(A|R)p(R) P(R) = p(L) = 1/4 Since the youngest person is already seated, there are only 4 places the next in line can sit.

Now, p(A|L) = probability(the third youngest sits next to the second, the fourth next to the third and the fifth in the last seat) Again we could apply conditional probability

= p(3 next to 2 | L) p(4 next to 3| L, 3 next to 2) p(5 in last seat | 4 next to 3, 3 next to 2 , L)

p(3 next to 2 | L) = 1/3 since 2 seats are taken there are 3 places 3 can sit and only 1 of them is next to 2 Similarly p(4 next to 3| L, 3 next to 2) = 1/2 And

p(5 in last seat | 4 next to 3, 3 next to 2 , L)=1/1=1

So, p(A|L) = 1/3* 1/2* 1= 1/6

Since the problem is the same in both directions, By symmetry, p(A| R) = 1/6

So, subbing everything in, P(A) = 1/4* 1/6. + 1/4*1/6 = 1/12

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