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From point 2 I have marked ,

If the 4 persons are sitting in a round table, then they can be shifted four times. You can see by looking at figure a,b,c and d.

Now , I understand that by making ABCD sit in a row and changing their positions . It looks like their arrangements are different . Like CDAB , BCDA but also they have written only 4 orders for this. In real , isn’t it that there will be total 4! Arrangements for it and not 4.

Also , from the round table perspective. If you keeping Changing the angle you’re looking at the ABCD. You can see different arrangements but they are also same right. So their total arrangements should be equal to 1.

I didn’t understand for 2 ,

No of linear arrangements is n? I think it should be n!.

Also , how did we get 1/n in 4 and n! For 3.

S.M.T
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  • this might help https://math.stackexchange.com/questions/1104919/calculating-probability-people-in-a-row/1104950#1104950 – David P Feb 13 '21 at 05:05
  • @DavidP Didn’t understand why there can be 2 ways in which r people between A and B can be arranged – S.M.T Feb 13 '21 at 05:10
  • Because $A$ can be less than $B$ and vise versa. The arbitrary choice of what's "first" needs to be factored in – David P Feb 13 '21 at 05:13

1 Answers1

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You're thinking about it the right way - the explanation in the book is just not very clear. The point is: if we counted circular arrangements the same way we count linear arrangements, we'd get $n!$, but this would count each arrangement exactly $n$ times (once in each of its equivalent rotated forms, as illustrated by the figure on top), so the true count is $n!/n$.

For example, with 4 elements, there are 24 linear arrangements:

ABCD BCDA CDAB DABC

ABDC BDCA DCAB CABD

ACBD CBDA BDAC DACB

ACDB CDBA DBAC BACD

ADBC DBCA BCDA CADB

ADCB DCBA CBAD BADC

but only 6 circular arrangements. Each row above represents one of the circular arrangements (depicted in four different ways).

Karl
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  • Let’s say if n = 4. Then circular permutations = 4 and linear = 4!. (Also ,here r = 4). Now , 4!/4 = 3*2=6. So what does 6 mean ? – S.M.T Feb 13 '21 at 05:19
  • From your answer , I think you mean to say that n!/n is circular permutations but then why is it coming 6? – S.M.T Feb 13 '21 at 05:20
  • By counting each arrangement exactly n times ,Do you mean to say that if ABCD are in a round table.Please give an example ? – S.M.T Feb 13 '21 at 05:23
  • The figure in the book shows four versions of one circular arrangement. There are 6 different circular arrangements, and you could represent each one in four different ways as a linear arrangement. – Karl Feb 13 '21 at 05:25
  • Also , do you mean to say that ABCD = DABC=CDAC=BCDA as one thing since when we move around the table , it means same thing ? – S.M.T Feb 13 '21 at 06:57
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    That's right. In the book's terminology, those are different "linear arrangements" but the same "circular arrangement". The word "permutation" always means (something equivalent to) linear arrangement. – Karl Feb 13 '21 at 08:17
  • Thank you sir........ – S.M.T Feb 13 '21 at 08:50