mrblack wrote:
Is it E? Here is my reasoning:
Now let's find the # of combinations where there's a BGB combination. Let the sequence of BG=A:
BBAA
BAAB
BABA
ABBA
ABAB
AABB
So there are 6 combinations here. But since BG can also be GB, we have to multiply by 2. We have to multiply by another 2 because there are 2 As. Therefore, there are 6x2x2=24 combinations.
We also have to include the instances where GG occurs at the end or the front, that is:
GGBBBB
BBBBGG
(all the other instances such as BBGGBB have been taken care of in the calculation at the top)
So we have 24+2=26 possibilities of 4 boys and 2 girls.
The total number of possibilities is 2^6=64. Therefore, the probability is 26/64, which is 13/32. The answer is E, none of the above.
Your solution can be simplified this way as well:
Total number of events : 2^6 as each position can be filled in exactly two ways
Events in favor : (4 Boys and Two Girls; It is like 6 objects in which 4 are alike and 2 others are alike) so 6!/2!4!
If it were a probability question, I will follow
Bunuel's aproach but the solution above is also important if it is counting problem.
Question: If we have n objects out f which p are alike, q are alike, how many possible number of comnitations are possible ? If it were a permutation problem, answer is n!/p!*q! as in the question above