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War of the roses!! - Probability qn.

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War of the roses!! - Probability qn.

by Motherjane » Mon Jun 16, 2008 11:40 am
HI Guys,

The first post in the forum. Need help with the following question.

TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??

A] 1/5
B] 2/5
C] 3/5
D] 4/5
E] 1/20
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by chidcguy » Mon Jun 16, 2008 4:22 pm
Total number of ways 5 different people can be arranged is 5! ways

Total number of ways that no two couples are next to each other

=

Total number of ways 5 people can be arranged - Total number of ways two couples are next to each other all the time


Total number of ways two couples are next to each other all the time

= 3! X 2! X 2! (Imagine two couples as one unit 3!, 2! ways each couple can sit in between themselves )

120-24= 96

Ans = 96/120 = 4/5
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by pranavc » Mon Jun 16, 2008 8:18 pm
chidcguy wrote:

Total number of ways two couples are next to each other all the time

= 3! X 2! X 2! (Imagine two couples as one unit 3!, 2! ways each couple can sit in between themselves )
Hi,
I was wondering if you could be more explicit in establishing how the total number of ways two couples are next to each other all the time = 3! x 2! x 2!

I got the correct answer but I actually listed all possible combinations. I'll never end up finishing the Math section if I keep doing stuff like that so it really would be nice for me to understand your approach better. Thanks in advance.
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by Motherjane » Tue Jun 17, 2008 12:41 am
Thanks again for the post chidcguy!!

The answer is 2/5. Was just thinking aloud if we can look at the other way around.

I mean:
p(no couple together) = 1-p(atleast 1 couple together)

I haven't figured out the solution yet!!
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by chidcguy » Tue Jun 17, 2008 8:04 am
pranavc,

What I did is, to find the number of possibilities in which all the couples are together.

Bundle both the couples as one unit. 1 Single & 2 couples in 3! ways

Each couple can be arranged in 2! ways in between themselves

So thats 3! X 2! X 2! =24

So they are NOT together = 120-24 =96

Thats how I arrived at 96/120 = 4/5

How ever MJ says the answer is 2/5? What answer did you get?

Ian,

Can you point out where I went wrong?
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by durgesh79 » Tue Jun 17, 2008 8:32 am
chidcguy:

if X1Y1 and X2Y2 are two couple and S is the single person.

You are counting X1SY1X2Y2 in your answer of 4/5. As per question stem it should not be there.
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by Motherjane » Tue Jun 17, 2008 9:12 am
chidcguy wrote:pranavc,

What I did is, to find the number of possibilities in which all the couples are together.

Bundle both the couples as one unit. 1 Single & 2 couples in 3! ways

Each couple can be arranged in 2! ways in between themselves

So thats 3! X 2! X 2! =24

So they are NOT together = 120-24 =96

Thats how I arrived at 96/120 = 4/5

How ever MJ says the answer is 2/5? What answer did you get?

Ian,

Can you point out where I went wrong?

Thanks to Durgesh for pointing out the specific case where couples will be sitting with each other.

If we take those options where the couples are sitting with each other, then there will be two such instances as there are two couples.

If we take one couple as one unit, then there will be total 2! * 4! ways they can be arranged. 2! for the couple arranging among themselves.

Similarly for the other couple - 2! * 4! ways.
These two cases will include those cases mentioned by chidcguy.

So total such possibilities will be :
2! * 4! + 2! * 4! - 3! X 2! X 2! = 48+48-24 = 72

So the probability of at least one couple sitting with each other = 72/120 = 3/5

So the probability of no couples sitting with each other = 1-3/5 = 2/5

Please verify whether the approach is fine or not.
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by durgesh79 » Tue Jun 17, 2008 9:46 am
The approach is fine. this is the conventional method of doing it. Basically we are trying to find out.

P(AUB) = P(A) + P(B) - P(A IN B)

A = couple A together
B = couple B together

A IN B = Both A and B together
A U B = one of them together.

But personally i dont like this method. I dont have an alternate method but I'm waiting for Ian's reply. :) He always comes up with interesting ways.[/img]
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Re: War of the roses!! - Probability qn.

by Ian Stewart » Wed Jun 18, 2008 5:34 am
Thanks for the kind words :)

chidcguy's first solution answers a different question: 'what is the probability that at least one couple is separated'? We need both couples to be separated, of course. As Motherjane and durgesh point out above, we can use chidcguy's result and method to get the answer- it's an interesting solution to the problem, I think.
Motherjane wrote:TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??

A] 1/5
B] 2/5
C] 3/5
D] 4/5
E] 1/20
It's often easiest to break down the problem into cases- it's a technique we use all the time in counting problems. Let's call the single person S. S must be in the 1st, 2nd, 3rd, 4th or 5th seat:

-If S is in the first seat - 1 choice
-then anyone else can be in the 2nd seat: 4 choices
-in the 3rd seat, cannot be the partner of the person in the 2nd seat: 2 choices
-in the 4th seat: must be the partner of the person in the 2nd seat- 1 choice
-fifth seat: must be the partner of the person in the 3rd seat- 1 choice.

So if S is in the first seat, we have 1*4*2*1*1 = 8 arrangements. Notice we'll get the same answer when S is in the fifth seat.

It's just as fast to count the arrangements with S in the second/fourth seat (8 arrangements for each) and third seat (16 arrangements), which gives us 8+8+16+8+8 = 48 arrangements in total. So the answer must be 48/5! = 2/5.

There are many other ways to look at the problem, but to be honest, I'd llkely do the above case analysis on a real GMAT- it's sure to work, and it's quick enough. I don't see a "five second solution" to the problem, but there may well be one.
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by durgesh79 » Wed Jun 18, 2008 7:46 am
Simply excellent Ian !!!!!!
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by Motherjane » Wed Jun 18, 2008 2:53 pm
I concur with Durgesh!!

Excellent answer Ian.
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by kunsjoi » Sat Oct 25, 2008 11:11 pm
Great response Ian!! Mucho Thanks!!
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by dark_angel_2208 » Thu Dec 18, 2008 8:51 am
Sorry but I wonder a little bit about Ian's answer:

If S is in the 2nd/4th seat, why is there 8 ways? (in my opinion: 12 ways)
If S is in the 3rd seat, I count only 8 ways instead of 16 as you said
( I use the same method you suggested)

But my final answer is just the same because 8*2+12*2+8=48
Can you make it clearer? :wink:
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by GMATters1001 » Thu Dec 18, 2008 4:29 pm
Obviously, the best way to do this is to look at the possibility they will be sitting next to eachother subtracted from total 120 possibilities:

I did it this way:

Total # of ways: 120
LESS:
Total ways one of the two couples can be together: 2*4! (since each member of the couple can switch and there are 4 total units i.e. couple, 4 singles)
Total ways the other couple can be together: 2*4! (since each member of the couple can switch and there are 4 total units)
However, this double counts each time both couples are sitting next to eachother
ADD BACK:
The repeats, i.e. the number of ways both couples are sitting next to eachother since they were counted twice. Think of them as one unit so there are 3 total units arranged 3! ways and you get 2*2*3! since each member of two couples can switch.

Formula is 120-2*4!-2*4!+2*2*3! = 120-48-48+24 = 48

48/120 = 2/5

-or you could look at it as reducing the 2nd reduction for one couple sitting next to eachother by the times that the 2nd couple is also sitting next to eachother, since those combinations were already counted
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by Ian Stewart » Fri Dec 26, 2008 1:34 pm
dark_angel_2208 wrote:Sorry but I wonder a little bit about Ian's answer:

If S is in the 2nd/4th seat, why is there 8 ways? (in my opinion: 12 ways)
If S is in the 3rd seat, I count only 8 ways instead of 16 as you said
( I use the same method you suggested)

But my final answer is just the same because 8*2+12*2+8=48
Can you make it clearer? :wink:
Sure...

We'll get the same number of possibilities when S is in the second seat as when S is in the fourth. If S is in the second seat:

1st seat: any of the 4 people different from S
2nd seat: must be S
3rd seat: can *only* be one of the 2 people not married to the person in the first seat
4th seat: *must* be the one person married to the person in the first seat
5th seat: must be the remaining person

Notice that the couples absolutely must be in seats 1+4 and seats 3+5 here, and we have only 8 possibilities in total.

Finally, if the single person is in the third seat:

1st seat: any of the 4 people
2nd seat: any of the 2 people from the other couple
3rd seat: the single person
4th seat: any of the 2 remaining people
5th seat: the 1 remaining person

giving 16 possibilities in total.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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