Hi All,

We're told that six children - A, B, C, D, E, and F - are going to sit in six chairs in a row, but Child E must be somewhere to the LEFT of Child F. We're asked for the number of possible configurations of the children. This question is a permutation question with a small twist - and can be solved with permutation math and a little logic.

To start, if all 6 children could sit anywhere, then there would be 6! = (6)(5)(4)(3)(2)(1) = 720 possible arrangements.

However, Child E must be to the LEFT of Child F. This 'twist' is not as complicated as it might first appear though. When considering all 720 possibilities, each child has an equal likelihood of sitting in any individual chair. By extension, HALF the time Child E will be to the left of Child F (or any other Child, for that matter). Thus, we have to remove the other half of the possibilities (in which Child E is to the right of Child F).... 720/2 = 360

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Here are a few similar questions to practice with:
- https://www.beatthegmat.com/permutation-t261691.html
- https://www.beatthegmat.com/vowels-and-c ... 49946.html
- https://www.beatthegmat.com/how-many-pos ... 77574.html

Cheers,
Brent

BTGmoderatorDC wrote:Six children - A, B, C, D, E, and F - are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720

If there is no restriction, the number of seating arrangements is 6! = 720. Of these 720 arrangements, half of them will have E sitting to the left of F (and the other half will have E sitting to the right of F). Therefore, the number of possible seating arrangements with E is to the left of F is 360.

Answer: D