Problem Solving

Well while explaining you I mised out something that there are five chairs

You are perfectly correct.
Answer has to be 3/5

I don't really know the source of this question but since I belong to Engineering background, we are used to face probabilty questions of much higher difficulty level.

Moreover I have posted some methods for speed calculations, I will just check that out and post the link,it might be of some use to you.

Hope this helps you!

Karan

https://www.beatthegmat.com/speed-calcul ... 33641.html


This is the link for the method I posted for making fast calculations.

Check it out and let me know.

Karan

Actual answer is 2/5
this is from princeton review and here is explanation given by them.

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 × 2 × 1 × 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 × 2 × 1 × 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 × 2 × 2 × 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. Whew!

Hi, I have a slightly different approach -

Now lets call the couples 'a' 'a1' and 'b' 'b1' and the single person as 'c'

The position of 'c' will determine the permutations

1) If 'c' occupies the first seat:

C _ _ _ _

The second seat can be filled in 4 ways (a, a1, b, b1)
The third seat can bee filled in 2 ways (for eg. if 'a' filled the second seat, 'b' or 'b1' could be seated in the 3rd)
The fourth seat can be filled in 1 way (a1)
The fifth seat can be filled in 1 way (b1)

This way none of the couples sit next to each other.

Thus total permutations = 1 * 4 * 2 * 1 * 1 = 8ways

Now apply the same logic to the remaining 4 cases

2) If 'c' occupies second seat -

_ c _ _ _

4 * 1 * 2 * 1 * 1 = 8 ways

3) If 'c' occupies third seat -

_ _ c _ _

4 * 2 * 1 * 2 * 1 = 16 ways

4) If 'c' occupies fourth seat -

_ _ _ c _

4 * 2 * 1 * 1 * 1 = 8 ways

5) If 'c' occupies fifth seat -

_ _ _ _ c

4 * 2 * 1 * 1 * 1 = 8 ways

SO total ways in which they can be seated without any couple sitting next to each other = 8 + 8 + 16 + 8 + 8 = 48

Total ways in which they can be arranged = 5!

Thus the required probability = 48/120 = 2/5


Please let me know what is the OA

_ _ _ _ _
in the 5 seats above the single person can only sit in "2" ways i.e. at the ends to allow the four of them to sit together
so the single person sits in 2 ways and the couple can rearrange themselves in 4! ways so
the couple can sit together in 2 x 4! ways right
...?

divide this by the total ways - 5!
will result in 2/5

now 1- (2/5) as we found out the prob for couple being together
the resultjavascript:emoticon(':?:')javascript:emoticon(':roll:')javascript:emoticon(':|') being 3/5
where did i go wrong
please mention the error in my approach stated above
???

Tryingmybest 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

OA -[spoiler]D bit I get 2/15[/spoiler]

I am not able to understand how my calculation is not correct.

5 people = individual 1 (I1) + couple 1 (C1)+couple 2 (C2)

I will try to find possibilities where c1 and c2 sit together. In that case, we have 3 persons sitting =3!. Plus each set of couple can arrange among themselves in 2! ways. Therefore nos of ways in which 5 people can sit (with couples sitting adjacent) = 3! 2! 2! = 24

Probability of not sitting together = (1 - 24/5!) = 4/5

Can someone help me where I am wrong

confused!! help...

Vemuri wrote:

cm47323 wrote:It is easy to figure out #ways both couples are next to each other, but to figure #ways exactly 1 couple is next to each other is not apparent to me.

How should we solve if the question is saying that the couple should be seated together?

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Hi Vemuri, since both Brent Hanneson and mals24 have already given some excellent approaches to the main problem, so I would try my hand on P (1 - your query) only with a little to say to everyone.

We have A, A', B, B', and C to arrange along five chairs such that A, A', B, B', C kind of arrangements are never there.

Counting is a two-mouthed snake and we are hardly ever put right in the middle of this snake. Hence, we are by and large provided with a shorter exit to save life that is as important as saving time when it comes to counting.

The present problem is also testing us on identifying the shorter exit, which is like thinking otherwise, or first answering opposite of what is coming out to be extremely confusing and dicey, if taken as it is.

Let's first try to answer the probability that the couples sit together in adjacent chairs?

For this we can take A, A', B, B', C as a group of three, (A, A'), (B, B'), C; this can be permuted in 3! × 2! × 2! ways. Out of a total of 5! Possible arrangements, 3! × 2! × 2! are favoring otherwise, hence 5! - 3! × 2! × 2! = 96 ways would favor the chief query.

The required probability = 96/120 = 4/5

See, the probability doubles just because C is no more a bone of content to the couples and could enjoy the corner seats as well.

Brent Hanneson wrote:

gmat740 wrote:

Tryingmybest 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

OA -D bit I get 2/15


# of ways to sit the couple together: Glue the couple together. They are now one entity. Now remove a chair since this one entity still represents two people and, thus two chairs.
Seat the couple: 4 chairs --> 4 ways.
Seat the single person. 3 chairs remaining -->3 ways
The total number of ways to seat everyone is 4x3.
BUT there are two ways we can glue the couple together (MW or WM)
So, the total number of ways to seat everyone is 4x3x2=24

# of ways to sit everyone:
Seat one person. 5 chairs --> 5 ways
Seat next person. 4 chairs remaining --> 4 ways
Seat last person. 3 chairs --> 3 ways
Total = 5x4x3 = 60

Probability = 24/60 = 2/5

@ Brent:

What Have you calculated!!!

take a look at the bold face


The probability of couple sitting together is 24

So the number of ways of not sitting together will be
( total ways- ways in which they sit together)

60-24 = 36


So probablity is 36/60

= 3/5

Please check the OA

You're 100% correct.
That's the second question this weekend that I didn't read it through.
I calculated the probability that the couple IS together.
So, we need to subtract that from 1 to get 3/5

Why regret for an outstanding work, Brent? Did I miss a thing?

Vemuri wrote:Thank you for replying GMAT740. I have a retraced your explanation with a correction:

The question mentions only 5 chairs:

[C1] [C2] [] [] [] --> S can seat in 3 different ways.
[] [C1] [C2] [] [] --> S can seat in 3 different ways.
[] [] [C1] [C2] [] --> S can seat in 3 different ways.
[] [] [] [C1] [C2] --> S can seat in 3 different ways.

Hence, the arrangements = 4 X 3 X 2 = 24

Total possible arrangements is 5!/2! (the 2! is the arrangement of C1 & C2, i.e. C1C2 or C2C1) ==> 120/2 = 60

The probability of the couples sitting together is 24/60 = 2/5
Hence, the probability of the couples not sitting together is 1-2/5 = 3/5.

Is this the correct answer? TryingmyBest, can you please check the answer options again? What is the source of this question?

i too solve these problem...... for me also 3/5 is coming .......check it once .......what OA?........

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?

Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2

There are five seats and in all there are five people so we have 5! = 120 different combinations right!
Now there are two couples which can be arranged only together as

x x x x x
1 2 3 4 5
first couple can be arranged in 12 , 23, 34, 45, i.e in 4 ways
other couple can be arranged in remaining 3 ways
last person has no option as rest of 4 seats are occupied so only 1 way
But both of two couples can be arranged in 2 ways

so total no of ways with couple together is 4*3*1*2*2*= 48
Now desired probability is 1- 48/120= 3/5

Am i Right? please respond

Tryingmybest 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

OA -[spoiler]D bit I get 2/15[/spoiler]

We could treat this as an overlapping groups problem:

Total = G(1) + G(2) - Both + Neither

Let's say that we have couple AB, couple CD, and lonely person E.

Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)

We need to subtract from the arrangements in which both AB and CD sit together because when we count the arrangements in which AB sit together, among those arrangements will be some in which CD also sit together. When we count the arrangements in which CD sit together, among those arrangements will be some in which AB also sit together. So the arrangements in which both AB and CD sit together -- the overlap -- will have been counted twice.

Total arrangements = 5! = 120

Total arrangements with AB together means we're arranging the 4 elements AB, C, D, and E = 4! = 24. Now we multiply by 2 since AB can be reversed to BA: 4*24 = 48.

Total arrangements with CD together means we're arranging the 4 elements CD, A, B and E = 4! = 24. Now we multiply by 2 since CD can be reversed to DC: 2*24 = 48.

Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24.

Plugging into the overlapping groups formula, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

So P(neither couple sitting together) = 48/120 = 2/5.

The correct answer is D.

GMATGuruNY wrote:

Tryingmybest 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

OA -[spoiler]D bit I get 2/15[/spoiler]

We could treat this as an overlapping groups problem:

Total = G(1) + G(2) - Both + Neither

Let's say that we have couple AB, couple CD, and lonely person E.

Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)

We need to subtract from our total the arrangements in which both AB and CD sit together because when we count the arrangements in which AB sit together, among those arrangements will be some in which CD also sit together. When we count the arrangements in which CD sit together, among those arrangements will be some in which AB also sit together. So the arrangements in which both AB and CD sit together -- the overlap -- will have been counted twice.

Total arrangements = 5! = 120

Total arrangements with AB together means we're arranging the 4 elements AB, C, D, and E = 4! = 24. Now we multiply by 2 since AB can be reversed to BA: 4*24 = 48.

Total arrangements with CD together means we're arranging the 4 elements CD, A, B and E = 4! = 24. Now we multiply by 2 since CD can be reversed to DC: 2*24 = 48.

Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24.

Plugging into the overlapping groups formula, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

So P(neither couple sitting together) = 48/120 = 2/5.

Thanks GURU .. Great explanation

GMATGuruNY wrote:...Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24...

Shouldn't we multiply by 8 because of 3 reversals?

edge wrote:

GMATGuruNY wrote:...Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24...

Shouldn't we multiply by 8 because of 3 reversals?

The number of ways to arrange the 3 elements AB, CD and E = 3! = 6.
There are 4 variations in which AB and CD sit as couples: AB and CD, BA and CD, AB and DC, BA and DC.
Thus, we multiply by 4:
4 * 3! = 24.