In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?	Choose the correct choice from the following choices
A) 1,008	B) 1,296	C) 1,512	D) 2,016	E) 2,268
We want A, B, and C to be before F, G and H.   Let's deal with this constraint first: consider A, B, and C as one unit {ABC} and F, G and H also as one unit {FGH}. According to the condition we need them to be arranged as {ABC}{FGH}. A, B, and C, within their unit can be arranged in 3! ways, similarly F, G and H , within their unit can also be arranged in 3! ways.   So, we have a line with 6 people XXXXXX. D can take any place in the line so D has 7 options: *X*X*X*X*X*X*.   Now, we have a line with 7 people and E can take any place there, so E has 8 options: *X*X*X*X*X*X*X*.   Total = 3!*3!*7*8 = 2,016.   Answer: 2,016.
<result>D</result>
In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?