voodoo_child wrote:In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?
A 5760
B 14400
C 480
D 56
E 40320
Here's how I would set this up:
Take
11 chairs (yes 11), and first seat the 5 boys in chairs 2, 4, 6, 8, and 10
_B_B_B_B_B_
This can be accomplished
5! ways (i.e.,
120 ways).
Note: This arrangement prevents the girls from sitting together.
Now seat each of the 3 girls in one of the 6 remaining seats.
The first girl can sit in any of the
6 seats.
The second girl can sit in any of the
5 remaining seats.
The third girl can sit in any of the
4 remaining seats.
So, we can seat the three girls is (
6)(
5)(
4) ways (i.e.,
120 ways)
At this point, throw away the 3 empty seats, and you have 8 children seated.
So, the total number of ways to seat all of the boys and girls is (
120)(
120) = [spoiler]14400 = B[/spoiler]
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
