Three pairs of siblings, each pair consisting of one girl

By far the fastest and most elegant solution!

Those who want to play with combinatorics are invited to provide an alternate solution.

Thanks Bunuel, you once again showed that in GMAT in most cases it is more logical thinking than doing quants. I have tried this one with different approaches but still could not come up with solution, but after your explanation it seems so easy and i wonder how i could not come up myself.

Thanks!!!

Aren't there only 5! total arrangements around a table for 6 people?

We are not told that these 6 are seated around a table, so we don't have circular arrangement. The question implies that they are seated like in a row.

Hi, couldnt understand why to devide by 6! ??

Hi Bunuel,

first sibling can be seated in 1/2 ways. but how do we come about the second sibling probability of 1/2 ? I am bit confused here, can you explain please ?

thanks

Hi Eva I dont get eather of the solution , by Bunual or by you, please explain Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5

Let the siblings be : in the order
Girl Boy
A B
C D
E F

Let the arrangement be _ _ _ _ _ _

So if you place lets say A on the first dash (=6 ways you can place A), you only have 5 places to let B go to. Thus you get 6*5. Eva has divided this and other possible arrangements by 2 to account for the fact 50% of the combinations will have AB_ _ _ _ while 50% will be BA_ _ _ _ . Only cases with AB_ _ _ _ type of combinations are allowed. We can safley assume 50% for either cases as there is no case for a 'bias' in these arrangements.

Bunuel has done the same , albeit in a slightly different manner. Probability of any girl sibling sitting to the right of the boy sibling = 50% or 1/2 (same as above)

Final probability = probability of 1st sibling girl to the left of the boy sibling * probability of 2nd sibling girl to the left of the boy sibling *probability of 3rd sibling girl to the left of the boy sibling = 1/2 * 1/2 * 1/2 = 1/8

Probability can be calculated in 2 ways:

Probability = total favorable cases / total cases (which is what Eva has done) or

Probability = probability of case 1* probability of case 2*probability of case 3 etc .... (which is what Bunuel has done).

You can choose whichever method suits you.

Thanks Now I got it.little more help, where i am wrong in this ...


Let the siblings be : in the order
Girl Boy
A B
C D
E F

Only three cases are available

ABCDEF ....(3!) no of ways three siblings can be arranges like CDABEF(one of the case out of 6) = 6
+
ACEBDF ...3! *3! ( no of ways ACE and BDF can be arranged them self) = 36
+
ABCEDF OR CDAEBF OR EFACBD ( no of ways ne sibling comes extrem left , then remaining two girls then remaining two boys)

3c1(ne subling out of three) *2! (ways two girls arranged among themself ) * 2! (ways 2 boys arranged among themself ) =24

36+24+6/6!

Hi Bunuel,

Please let me know if my approach to the problem is correct. I have seen your other solutions where we multiply by 1/2 whenever we have a condition of sitting/ standing only on left or right.

6!*1/8 ==> 90 (favourable outcome) /6! ==> 90/720 ==> 1/8.

So instead of multiplying 1/2 for each pair, i took (1/2)^3, 3 is the number of pairs. I want to know if this approach is correct in case i have to apply to similar questions.

Let’s call one pair of siblings Abby and John. Since two people can be arranged in 2 ways, the probability is 1/2 that Abby sits to the left of John and 1/2 that she sits to the right of John.
Using that same logic, we see there is a 1/2 x 1/2 x 1/2 = 1/8 probability that all 3 girls are seated to the left of their male siblings.

Answer: D

I think Bunuel just did a beautiful job.
However, my solution is a little bit quantitative.

Let's call L the left position of the girl, and R the right one.
The question is: what is the probability that L-L-L happens ?
The following cases would happen:
L-L-L : 1 case
R-R-R : 1 case
L-R-R : 3 cases (choose 2 out 3 pairs that have the girl sitting on the right)
L-L-R: 3 cases (choose 2 out 3 pairs that have the girl sitting on the left)
Total cases: 8
The probability = 1/8.

Probability of being on the right is equal to probability of being on the left so for each pair the probablity is 1/2. Therefore, the ans would be : (1/2)^3 = 1/8

Posted from my mobile device

Pick the first male, 3 choices.

Only 1 choice for his sibling 🤪.

One choice for which side to place the sibling, so a total of:

3*1*1=3 choices

Place 2nd male, two choices.

1 choice for sibling, 1 choice, left, for seat.

Total choices:

2*1*1= 2

3rd sibling pair only 1 choice


Total choices meeting requirements:

3*2*1= 6

Total available choices follow logic above, but recognizing that female sibling has 2 choices of left or right for each pair,so

3*2=6
2*2=4
1*2=2

Or 6*4*2= 48 potential arrangements.

Probability: 6/48= 1/8


Note that there is no reduction in available seats to the left or right just because a prior pair has been seated, since there is no limitation in the question to the number of seats, so one can assume an equal availability of seats to the left or right or 1/2 probability of being seated to the left or right for all sibling pairs. This is based on an infinite number of seats point of view.

The other point of view is that even with a limited set of seats, every pair has the same opportunity to be properly seated since no specific pair has been identified as occupying a specific set of seats, thereby not constraining the remaining pairs choices.

Posted from my mobile device

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.