it isnt tough but i still didnt get it

sana.noor 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

OA is B

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_

Since we are arranging 5 unique things (boys), this step 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