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by voodoo_child » Mon Sep 24, 2012 12:21 pm
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 what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?

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by Brent@GMATPrepNow » Mon Sep 24, 2012 2:05 pm
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
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