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

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07 Jan 2015, 09:21

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

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18 Jan 2016, 07:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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26 Feb 2016, 15:35

Bunuel wrote:

arnaudl wrote:

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs? A: 1/5 B: 1/4 C: 3/8 D: 2/5 correct E: 1/2

Ignoring the constraint first : total number of arrangements = 5! = 120 Then, I try to consider the nb of arrangment where at least one couple sits together but I cannot find using formulas the 72 required. (ie the 3/5 required which leads to the correct answer choice 1 - 3/5 = 2/5) ?

Help please !

Let's find the opposite probability and subtract it from 1.

Opposite event that neither of the couples sits together is event that at leas one couple sits together. # of arrangements when at leas one couple sits together is sum of arrangements when EXACTLY 2 couples sit together and EXACTLY 1 couples sit together.

Couple A: A1, A2 Couple B: B1, B2 Single person: S

EXACTLY 2 couples sit together: Consider each couple as one unit: {A1A2}{B1B2}{S}, # of arrangement would be: \(3!*2!*2!=24\). 3! # of different arrangement of these 3 units, 2! arrangement of couple A (A1A2 or A2A1), 2! arrangement of couple B (B1B2 or B2B1).

EXACTLY 1 couple sits together: Couple A sits together: {A1A2}{B1}{B2}{S}, # of arrangement would be: \(4!*2!=48\). 4! # of different arrangement of these 4 units, 2! arrangement of couple A (A1A2 or A2A1). But these 48 arrangements will also include arrangements when 2 couples sit together, so total for couple A would be \(48-24=24\);

The same for couple B: {B1B2}{A1}{A2}{S}, # of arrangement would be: \(4!*2!=48\). Again these 48 arrangements will also include arrangements when 2 couples sit together, so total for couple B would be \(48-24=24\);

\(24+24=48\).

Finally we get the # of arrangements when at least one couple sits together is \(24+48=72\).

Total # of arrangements of 5 people is \(5!=120\), hence probability of an event that at leas one couple sits together would be \(\frac{72}{120}=\frac{3}{5}\).

So probability of an event that neither of the couples sits together would be \(1-\frac{3}{5}=\frac{2}{5}\)

Answer: D.

Hope it's clear.

How did you get to be so good at Math?
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Re: Two couples and one single person are seated at random in a [#permalink]

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29 Aug 2016, 06:19

arnaudl wrote:

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5 B. 1/4 C. 3/8 D. 2/5 E. 1/2

Hi All

Can't we solve this problem like this: (A1,A2,B1,B2,C)

Since no one sits together and we have 5 chairs, i.e. we can arrange people like: _ X _ X _ (where X can be anyone person from the 2 couples) No of ways of selecting who sits on 2nd chair: 4C1 = 4 No of ways of selecting who sits on 4th chair: 1 (it has to be the other person from couple) Rest 3 places can be filled in 3! ways = 6 Total ways of arranging 5 people = 5! = 120 Answer = (4*6)/120 = 1/5

Two couples and one single person occupy a row of five chairs at rando [#permalink]

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25 Jul 2017, 07:16

Two couples and one single person occupy a row of five chairs at random. What is the probability that neither couple sits together (the husband and the wife should not occupy adjacent seats)?

Two couples and one single person occupy a row of five chairs at random. What is the probability that neither couple sits together (the husband and the wife should not occupy adjacent seats)?

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

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13 Aug 2017, 03:25

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.

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