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Two couples and one single person are seated at random in a

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Manager
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B
Joined: 03 May 2017
Posts: 113

Kudos [?]: 19 [0], given: 14

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Re: Two couples and one single person are seated at random in a [#permalink]

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New post 26 Jul 2017, 01:59
An approach: ways for couples to sit
Aa/Bb/C = 4C2* 2*2*2= 48 ways
Now Take each couple as an entity i.e Aa=A and Bb=B
A/B/C= 3C2* 2*2*2= 24 ways

Total ways to sit together = 24+48= 72
Probability of not sitting together = (5!-72)/5! = 2/5.

Bunuel, thoughts on this?

Kudos [?]: 19 [0], given: 14

Manager
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Joined: 06 Jul 2014
Posts: 104

Kudos [?]: 40 [0], given: 179

Location: India
Concentration: Finance, Entrepreneurship
GMAT 1: 640 Q48 V30
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Re: Two couples and one single person are seated at random in a [#permalink]

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New post 13 Aug 2017, 02:25
Let's call the first couple C and c, the second couple K and k, and the single person S.
Let's seat S in different places and figure out the possible ways to have no couples sit together.
If S sits in the first seat, any of the remaining four people could sit next to S.
However, only two people could sit in the next seat:
the two who don't form a couple with the person just seated).

For example, if we have S K so far, C or c must sit in the third seat.
Similarly, we have only one choice for the fourth seat:
the remaining person who does not form a couple with the person in the third seat.
Because we have seated four people already, there is only one choice for the fifth seat;
the number of ways is 4 X 2 X 1 X 1 = 8.
Because of symmetry, there are also 8 ways if S sits in the fifth seat.

Now let's put S in the second seat.
Any of the remaining four could sit in the first seat.
It may appear that any of the remaining three could sit in the third seat, but we have to be careful not to leave a couple for seats four and five.
For example, if we have C S c so far, K and k must sit together, which we don't want.
So there are only two possibilities for the third seat.
As above, there is only one choice each for the fourth and fifth seats.
Therefore, the number of ways is 4 X 2 X 1 X 1 = 8.
Because of symmetry, there are also 8 ways if S sits in the fourth seat.


This brings us to S in the third seat.
Any of the remaining four can sit in the first seat.
Two people could sit in the second seat (again, the two who don't form a couple with the person in the first seat).
Once we get to the fourth seat, there are no restrictions.
We have two choices for the fourth seat and one choice remaining for the fifth seat.
Therefore, the number of ways is 4 X 2 X 2 X 1 = 16.
We have found a total of 8 + 8 + 8 + 8 + 16 = 48 ways to seat the five people with no couples together;

there is an overall total of 5! = 120 ways to seat the five people, so the probability is = 2/5.

Kudos [?]: 40 [0], given: 179

Re: Two couples and one single person are seated at random in a   [#permalink] 13 Aug 2017, 02:25

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