Combinations

[raj]

Source: Princeton GMAT Practice test

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

What should be the easiest strategy and what level of question do u think this is ?

Thanks

[RonPurewal]

since this problem deals with a relatively small # of people and is too complicated to be treated with a simple strategy (like an 'anagram grid'), my vote goes to organized counting.

[public service announcement]
if you decide quickly to use organized counting, you'll have plenty of time to do it. if you stare at the problem for a minute or more before deciding that you should use organized counting, you won't have time. make these decisions quickly; if you don't see an elegant solution right away, then start looking for the 'grunt' solution.
[/public service announcement]

let's say the couples are a-b and x-y.

here are all the ways a-b can sit in the chairs with 'a' listed first.
a_b__ (--> x-y could sit in 2-4 or 2-5)
a__b_ (--> x-y could sit in 2-5 or 3-5)
a___b (--> x-y could sit in 2-4)
_a_b_ (--> x-y could sit in 1-3, 1-5, or 3-5)
_a__b (--> x-y could sit in 1-3 or 1-4)
__a_b (--> x-y could sit in 1-4 or 2-4)

that list contains 12 possibilities.
since you could switch a-b or x-y, double this number twice, so there are 48 possibilities total.

there are 5! = 120 ways to arrange the five people altogether.

48/120 = 2/5

--

this problem should be considered difficult, as (1) probability problems are usually considered difficult by default, and (2) this one is tougher than the average probability bear.

[shaji]

Source: Princeton GMAT Practice test

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

What should be the easiest strategy and what level of question do u think this is ?

Thanks

The quick fix for this problem is to consider the three possibilities from the couples point of view namely:
1)Both couples sit together.
2)Only one couple sits together.
3)None of the couples sit together.

Now the pattern of the probability for such situations is unique and symetric, i.e. Probability(Prob) of 2=(Prob) of 3.*
Notice (Prob) 1+2+3=1
Therefore its very easy to compute Prob 3 as 0.4
* To emphasize this principle,consider the situation when there are only 2 couples without that single person!!!
Prob1=Prob2=Prob3=1/3.

This approach will take hardly a minute saving you over 70% time to focus on problems that may need more time.

The GMAT ,as the name suggests, is all about testing your managerial aptitude and 'quick fixes' that is great skill set that is greatly rewarded on the GMAT. I state this from my experience in Management School and my experience in management functions.

[mrinalini]

Ron I dont see how there are 48 possibilities here could you please explain this a bit more? Thanks.

[Guest]

Tough Problem.

Shaji's solution appeared quick, but I couldn't quite understand it.

The couples are AB and XY.

Consider the arrangements where A and B are sitting together.

A B _ _ _

_ A B _ _

_ _ A B _

_ _ _ A B

For each of the four cases, there are 3! ways (6 ways) the other people could be arranged, making for a total of 24 ways that A and B could be next to each other. If you reverse A and B, you get 48 ways.

Do the same thing for X and Y, and you get 96 total ways.

BUT!

Look at the first six cases, using the model above

ABXYZ, ABZYX, ABYXZ, ABYZX, ABZXY, ABXZY...

Half of these arrangements are redundant; they will be exactly the same when you are accounting for the possibilties where Y and X are together.

So, 96/2 = 48.

This would probably take too long on the test, but it does exhibit the conceptual foundation behind the problem.

[Guest]

The previous solution is not quite correct. Out of the first six cases, four are redundant. Out of the next six cases, two are redundant (ZABYX, ZABXY). So, out of the first twelve cases, six are redundant. The same is true of the next twelve cases, and of each twelve cases. But it is true that half are redundant.

[RonPurewal]

The quick fix for this problem is to consider the three possibilities from the couples point of view namely:
1)Both couples sit together.
2)Only one couple sits together.
3)None of the couples sit together.

Now the pattern of the probability for such situations is unique and symetric, i.e. Probability(Prob) of 2=(Prob) of 3.*
Notice (Prob) 1+2+3=1
Therefore its very easy to compute Prob 3 as 0.4

how do you arrive at prob(3) = 0.4?

from what you've written here, you conclude that p(2) = p(3) = x and p(1) = y. but that still gives 2x + y = 1, which admits a whole ton of possibilities; you seem to have randomly selected x = 0.4 out of thin air. why not x = 0.3 and y = 0.4? or 0.2 and 0.6? or 1/3 and 1/3?

--

also:

you might be right about the symmetry between cases 2 and 3 (i didn't go back and calculate case 2, but i suppose i could if you want), but, if those two cases do have the same probability, then they do only by coincidence. there's no true symmetry in the problem - you can't do something like switch the couples around to match the possibilities 1:1.

consider something mundane like flipping three coins. using the same type of logic, one might make the (mistaken) declaration that getting one head is just as likely as getting no heads.

be very wary of assigning symmetry to situations in which the symmetry is not totally obvious.

* To emphasize this principle,consider the situation when there are only 2 couples without that single person!!!
Prob1=Prob2=Prob3=1/3.

that does turn out to be true. but is there an obvious way to see that that must be so?

if so, spill!

The GMAT ,as the name suggests, is all about testing your managerial aptitude and 'quick fixes' that is great skill set that is greatly rewarded on the GMAT.

you are absolutely right about the quick fixes: the gmat is, above almost all else, a test of fast reasoning. many potential high scorers crash and burn because of poor time management.

i don't agree about managerial aptitude; the gmat is purely a test of lightning quick reasoning ability. there are plenty of whiz kids out there who could swoop in and score 750-800 cold on the gmat, but couldn't manage other people if their lives depended on it. there are also plenty of individuals who would naturally make great managers, but who have lots of trouble with the specific kind of reasoning tested on the gmat. that's why our company exists. :)

but i digress.

[RonPurewal]

Ron I dont see how there are 48 possibilities here could you please explain this a bit more? Thanks.

first, look at the list of cases. there are 12 different arrangements of the couples; go count them up.

but each of those 12 arrangements corresponds not to one possibility, but to four. i'll use an illustration to show you why:
the very first case listed is a-b in seats 1 and 3, and x-y in seats 2 and 4.
because you can switch people and get an arrangement that's genuinely different, that actually encompasses four possibilities:
axby_
aybx_
bxay_
byax_
same reasoning works for every other possibility listed, so you have to multiply the 12 possibilities by 4 (yielding 48)

[shaji]

The quick fix for this problem is to consider the three possibilities from the couples point of view namely:
1)Both couples sit together.
2)Only one couple sits together.
3)None of the couples sit together.

Now the pattern of the probability for such situations is unique and symetric, i.e. Probability(Prob) of 2=(Prob) of 3.*
Notice (Prob) 1+2+3=1
Therefore its very easy to compute Prob 3 as 0.4

how do you arrive at prob(3) = 0.4?

from what you've written here, you conclude that p(2) = p(3) = x and p(1) = y. but that still gives 2x + y = 1, which admits a whole ton of possibilities; you seem to have randomly selected x = 0.4 out of thin air. why not x = 0.3 and y = 0.4? or 0.2 and 0.6? or 1/3 and 1/3?

--

also:

you might be right about the symmetry between cases 2 and 3 (i didn't go back and calculate case 2, but i suppose i could if you want), but, if those two cases do have the same probability, then they do only by coincidence. there's no true symmetry in the problem - you can't do something like switch the couples around to match the possibilities 1:1.

consider something mundane like flipping three coins. using the same type of logic, one might make the (mistaken) declaration that getting one head is just as likely as getting no heads.

be very wary of assigning symmetry to situations in which the symmetry is not totally obvious.

* To emphasize this principle,consider the situation when there are only 2 couples without that single person!!!
Prob1=Prob2=Prob3=1/3.

that does turn out to be true. but is there an obvious way to see that that must be so?

if so, spill!

The GMAT ,as the name suggests, is all about testing your managerial aptitude and 'quick fixes' that is great skill set that is greatly rewarded on the GMAT.

you are absolutely right about the quick fixes: the gmat is, above almost all else, a test of fast reasoning. many potential high scorers crash and burn because of poor time management.

i don't agree about managerial aptitude; the gmat is purely a test of lightning quick reasoning ability. there are plenty of whiz kids out there who could swoop in and score 750-800 cold on the gmat, but couldn't manage other people if their lives depended on it. there are also plenty of individuals who would naturally make great managers, but who have lots of trouble with the specific kind of reasoning tested on the gmat. that's why our company exists. :)

but i digress.

Interesting comments!!! No doubt, Manhattan Gmat exists for all the right reasons and I am sure many will join me in appreciating what U do.

Certainly there is a genuine confusion with my interpretation and Maths , the queen of all sciences,will tolerate nothing short of perfect logic and rationale or perhaps 'spills'.

Now here with the 'spills':
1) p(3)=0.4;
Yes , the 'tons of possibilities" for p(3) is actually one and only one!!! if you consider p(1), which is most orderly situation(the least likely to happen) among the three possibilities that can occur.
The number of ways p(1) can occur are 2*2*3!=24.Therefore p(1)=24/120=1/5( the choice among the possibile answers 'thin air'. Remember, orderliness is least likely event in all randomn phenomenon.

2)In Mathematics there is no 'coincidence' ,only rationale. U have already noticed the symmetry when there are only two couples. Consider 3 couples and one single person and U will see the fun!!!. Further, the consideration of coins is irrelevant in this situation as coins are indistinguishable while the couples are. This is a typical case of extrasensory deception, a phenomenon sometimes seen is such standardized tests.

Finally; I agree some 'kids might manage a 'cold 750'' and may not have management aptitude. This could be a deficiency in the test rather than the 'kids' I am sure as others the Management schools knows it pretty well and the GMAT is adapting to these 'challenges'

[samir.najeeb]

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 72/120=3/5.

So probability of an event that neither of the couples sits together would be 1-3/5=2/5

Answer: D.

[jnelson0612]

Excellent discussion everyone! Thank you for posting.