Rahul@gurome wrote:Solution:
Let people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be A, B, C, D, E, F, G and H respectively.
Now A, B and C must be before F,G and H.
Two cases can arise.
Either (1) D and E are together
or
(2) D and E are not together.
Consider (1) first.
So if A B C F G H are arranged like this and D and E have to be together, there are 7 places in which they can be put in 7P1 *2 or 14 ways.
Also A. B and C can be arranged among themselves in 3! or 6 ways and F, G and H can be arranged among themselves in 3! or 6 ways.
Or If D and E have to be together, they can be arranged in 14*6*6 = 504 ways.
Consider (2) then.
If they are arranged like this A B C F G H, there are 7 places between them where D and E can be placed.
This can be done in 7P2 or 7*6 = 42 ways.
Also A. B and C can be arranged among themselves in 3! or 6 ways and F, G and H can be arranged among themselves in 3! or 6 ways.
So the possible arrangements we can have is 42*6*6 = 1,512.
Or total number of arrangements is 1512+504 = 2016 ways.
The correct answer is (D).
Rahul..Could you please explain little bit how did you come up with
if A B C F G H are arranged like this and D and E have to be together, there are 7 places in which they can be put in 7P1 *2 or 14 ways.
and
If they are arranged like this A B C F G H, there are 7 places between them where D and E can be placed.
This can be done in 7P2 or 7*6 = 42 ways.
