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In how many ways can 6 people be seated at a round table if

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In how many ways can 6 people be seated at a round table if [#permalink] New post 15 Oct 2006, 08:32
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A
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C
D
E

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In how many ways can 6 people be seated at a round table if one of those seated cannot sit next to 2 of the other 5?

a) 720
b) 120
c) 108
d) 84
e) 48
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 [#permalink] New post 15 Oct 2006, 10:12
6 People in a round table can be seated in (6 - 1) ! ways = 120.

Now we need to subtract the number of cases when one of those is sitting next to 2 of the other 5.

We can consider as if 5 people are sitting in a row because it is round table.
Again consider 3 people, those who can not sit together, as a single unit –

So the possible arrangements among remaining people 5 – 3 + 1 Unit are = 3 !
And the 3 people unit can arrange among themselves in 3 ! ways.

So the possible cases when one of those is sitting next to 2 of the other 5 = 3 ! * 3 ! = 36

Total possible cases = 120 -36 = 84
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 [#permalink] New post 15 Oct 2006, 11:10
Can someone tell me where I go wrong reasoning it this way.

Say the six people are A B C D E F and A cannot sit next to E and F

There would be 6 ways to sit A, 3 ways to sit the second person (on A's rightside), 2 ways to sit the third person (on A's leftside), 3 ways to sit the fourth, 2 ways to sit the fifth, and 1 way to sit the sixth.

6x3x3x2x2x1 = 216

I get 216. Where am I going wrong using this approach???? :x
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 [#permalink] New post 15 Oct 2006, 13:46
anindyat wrote:
6 People in a round table can be seated in (6 - 1) ! ways = 120.

Now we need to subtract the number of cases when one of those is sitting next to 2 of the other 5.

We can consider as if 5 people are sitting in a row because it is round table.
Again consider 3 people, those who can not sit together, as a single unit –

So the possible arrangements among remaining people 5 – 3 + 1 Unit are = 3 !
And the 3 people unit can arrange among themselves in 3 ! ways.

So the possible cases when one of those is sitting next to 2 of the other 5 = 3 ! * 3 ! = 36

Total possible cases = 120 -36 = 84


I understand that we need the total number of ways minus the exceptions...

but could someone explain why seating 6 people is not 6*5*4*3*2*1 = 720?

Why is it (6-1)!
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 [#permalink] New post 15 Oct 2006, 21:09
I dont get the question.

how can 1 one of them not sit next to the other 2 in a round table?
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Re: PS : Combinatorics : seats around a table [#permalink] New post 15 Oct 2006, 23:31
n people can be seated around a round table in (n-1)! ways.

ok...let's find the no of ways in which that person is always seated next to 2 particular people.these 3 can be seated in 2 ways because the cetre position is fixed.
now we have a total of 3+1 people...note that 1 represents the group of those 3 people.
so 4 can be seated in (4-1)! ways = 6 ways.
hence total ways when 2 particular people are always next to one particular of them = 2*6=12 ways..

and total no of ways in which 6 people can be seated =(6-1)!=120 ways..

hence answer= 120=12 = 108 ways.

choice b as per me.
what's the OA?



londonluddite wrote:
In how many ways can 6 people be seated at a round table if one of those seated cannot sit next to 2 of the other 5?

a) 720
b) 120
c) 108
d) 84
e) 48
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 [#permalink] New post 16 Oct 2006, 10:19
OA is C.

OE: 6 people can be seated round around a table in 5! ways (would appreciate someones clarification on whether this is correct and why). There are 2 ways the two unwelcome guests could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from 5! giving a result of 108.

Clear as mud :-D

Edit : AK why can n people be seated in (n-1)! ways and not n!

Thanks
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 [#permalink] New post 16 Oct 2006, 10:24
London .... if you allow me to answer ur question.....

it is because we have to reserve one place (the one in n-1)

as a refrence from which we start arranging and counting (because in a round shape we have no end and start points

Hope this Helps
  [#permalink] 16 Oct 2006, 10:24
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