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Re: In how many ways can 6 people be seated at a round table if [#permalink]

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15 Oct 2006, 10:12

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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 â€“

So the possible arrangements among remaining people 5 â€“ 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

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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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????

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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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 â€“

So the possible arrangements among remaining people 5 â€“ 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?

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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15 Oct 2006, 23:31

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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?

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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16 Oct 2006, 10:19

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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

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

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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Re: In how many ways can 6 people be seated at a round table if [#permalink]

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21 Jan 2015, 02:39

I may not able to get the question properly. Let me put forward my thought process:

Required no. of ways = Total no. of ways - when one of them is sitting next to 2 of other 5.

Now this question statement i.e. "one of those seated cannot sit next to 2 of the other 5" seems tricky to me. Let me split in parts:

1. "One of those.." Who out of 6 ?? There can be 6 ways to choose one out of 6.

2. "2 of the other 5"..Who 2 out of 5?? There can be 10 ways to choose 2 out of 5.

3. "when one of them is sitting next to 2 of other" - Suppose B is one out of 6 and A & C are 2 out of 5. Then does statement means that B should be in mid of A and C?

I was little confused here. But then i understand yes it should mean B is in mid of A & C. Any comments here most welcome..

So now lets find the actual no of ways for each type:

No. of ways When AB are together = Now we are left with 4 + 1(AB) = 5 people. SO no of circular ways = (5-1)! = 4! * 2!( A and B can be arranged themselves)

Now bind ABC together we are left with 3 + 1(ABC) = 4 people. No. of ways it can be arranged is (4-1)! * 2!(as A and C can be interchanged) = 12 ways

Total no. of ways = (6-1)! = 5! So required no of ways = 5!- 3!*2! = 120 - 12 = 108. Now shall we not consider the above point 1 and 2? if yes, then we should multiply by 6 and 10, isn't??

So total ways = 6*10*108 ??

Let me tweak the question little bit(as this was the reason of my initial confusion). What if instead of "one of those seated cannot sit next to 2 of the other 5"

question says "one of those seated cannot sit next any 2 of the other 5". Then to find this i had below analysis.

Then to find out no of ways "when one of them is sitting next to any 2 of other" = No. of ways When AB are together + No of ways when BC are together - No . of ways

when ABC are together and position of A, B and C is fixed in that arrangement i.e. A,B,C are binded together but should not be arraneged themselves.

The reason we are subtracting "No . of ways when ABC are together and position of A, B and C is fixed in that arrangement." is we have counted this twice when we have summed up "No. of ways When AB are together" and "No of

ways when BC are together".

For example: No. of ways When AB are together - Includes.... ABC and No. of ways When BC are together also includes... ABC Given position of A, B and C is fixed.

So No. of ways When AB are together means 4 + 1 (AB) = 5 people to be arranged in circular way. So it should be (5-1)! * 2(A and B can be interchanged) = 48 ways.

Likewise, No of ways when BC are together = 48.

No . of ways when ABC are together but arranged themselves = 3 + 1 (A,B,C) = (4-1)! = 6 ways

The no of ways "when one of them is sitting next to any 2 of other 5" = 48 + 48 - 6 = 90.

Required no. of ways = Total no. of ways - when one of them is sitting next to any 2 of other 5. Required no. of ways = 120 - 90 = 30 ways.

Now again considering point 1 and 2 = 6*10*30 = 1800 isn't?

Hi, Can i solve this by taking total number of arrangements i.e 5!=120 ways and then subtract the restriction?

120-(arrangements B should not sit next to A and/or C) ?

is that a correct approach?

GMATinsight wrote:

Quote:

In how many ways can 6 people A,B,C,D,E,F be seated at a round table if B cannot sit next to A and/or C?

a) 720 b) 120 c) 108 d) 84 e) 36

Have Modified the Language to make it clearer

6 people are A, B, C, D, E and F and B can not sit next to A and C

Considering the Position of B is fixed,

We have to make sure that 2 person who sit next to B are out of D, E and F

i.e. No. of ways of choosing the neighbours of B = 3C2 = 3

and the no. of ways the Selected neighbours can arrange at the two position adjacent to B = 2!

i.e. The ways the B and The neighbours can be arranged = 3C2 *2! = 3*2 = 6

Now The no. of ways in which Remaining Three Individuals can be arranged on remaining 3 seats = 3!

Total Ways of making Six person seated such that B doesn't sit next to A and C = 3C2 *2!*3! = 36

That would be fine but a difficult approach as you will have to calculate three cases

Case-1: When A sits next to B and C does not sit next to B A can sit next to B in 2 ways (On B's right or B's left side) The next adjacent place of B can be occupied in 3 ways because C can't sit next to B Remaining three can sit in 3! ways So total ways = 2*3*3! = 36 ways

Case-2: When C sits next to B and A does not sit next to B C can sit next to B in 2 ways (On B's right or B's left side) The next adjacent place of B can be occupied in 3 ways because A can't sit next to B Remaining three can sit in 3! ways So total ways = 2*3*3! = 36 ways

Case-3: When A and C both sit on either sides of B A and C can sit in 2! ways on two places adjacent to B Remaining three can sit in 3! ways So total ways = 2!*3! = 12 ways

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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04 Jul 2015, 10:52

6 People Sitting Around a Round Table Without Any Restriction = (6-1)! = 5! = 120

Restriction = 1 person cannot sit around other two particular people Complement Condition = 3 People Will Always Sit together Now considering 3 People as one group along with other 3 people , total number of ways they can sit = (4-1) = 3! = 6 Ways But group of 3 Can also Adjust it self in 3! ways = 6 Ways Total Complement Ways = 6+6 = 12

Total Ways = 120 - 12 = 108

Please correct, if this is not the right way of solving the question

Re: In how many ways can 6 people be seated at a round table if [#permalink]

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20 Aug 2015, 03:48

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I knew the (n-1)! formula which doesn't solve the entire problem here so I shifted gears and came to the following approach.

You have 6 spots on the table. Let's imagine to fix the guy who can't sit with all the other ones on one of the spots. (1) As he can't stand 2 out of total of 5 people, we have 3 options for the 2 seats next to him - so 3C2 which equals 3. (2) For the remaining 3 spots we have 3! or 6 options. Multiply (1) and (2) and you get 18 seating arrangements.

But bear with me, we have to remember that we fixed the bad guy on just one spot. He can sit on every seat on the table, namely - 6. So multiply the 18 seating arrangements with 6 and you get 108 (C) _________________

Thank you very much for reading this post till the end! Kudos?

In how many ways can 6 people be seated at a round table if [#permalink]

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20 Aug 2015, 04:40

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

I get a different answer. Let me explain how I worked out.

I fixed the person say A, who cannot sit with 2 of the other 5, in one place. Now, the remaining 3 persons who can be seated to the next seats of A can be done in 3p2 ways (6 ways) Note that we don't need to select those 3 from 5 because when you keep the 2 away the remaining 3 can be seated with A. Now, the remaining 3 persons can be seated in the remaining 3 seats in 3p3 or 6 ways. Therefore, 6*6=36 ways.

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