BTGmoderatorDC wrote:Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600
B. 720
C. 1440
D. 4320
E. 4800
Since A and B must occupy adjacent positions, consider AB a single element in the arrangement.
The number of ways to arrange the 6 elements AB, C, D, E, F and G = 6! = 720.
In 1/2 of these arrangements, C will be to the LEFT of AB.
In the remaining 1/2 of these arrangements,C will be to the RIGHT of AB.
Thus, the number of arrangements in which C is to the right of AB = (1/2)(720).
Since AB can switch to BA -- doubling the total number of possible arrangements -- we multiply by 2:
(2)(1/2)(720) = 720.
The correct answer is
B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
