P&C and Probability

[TanmayShah]

There 8 boys and 3 girls. They are to be seated in line for a photograph. Find the probability that no two girls are seated together.

Can someone please help and solve the above problem?
What would a general approach for such problems?

[Anju@Gurome]

My sincere apologies.
As Brent pointed out in his post below my approach did considered some cases more than once.

Thanks Brent.

[Brent@GMATPrepNow]

There 8 boys and 3 girls. They are to be seated in line for a photograph. Find the probability that no two girls are seated together.

This is a very difficult problem that is probably beyond the scope of the GMAT.
I believe Anju's method counts a few arrangement more than once (although she may have since changed her solution).


Here's another way to set this up.

P(no girls together) = [# of ways to seat everyone with no girls together]/[# of ways to seat everyone]

# of ways to seat everyone
There are 11 children, so we can arrange them in 11! ways.

# of ways to seat everyone with no girls together
Take 17 chairs (yes 17), and first seat the 8 boys in chairs 2, 4, 6, 8, 10, 12, 14, and 16
_B_B_B_B_B_B_B_B_
This can be accomplished 8! ways

Note: This arrangement prevents the girls from sitting together.

Now seat each of the 3 girls in one of the 9 remaining seats.
The first girl can sit in any of the 9 seats.
The second girl can sit in any of the 8 remaining seats.
The third girl can sit in any of the 7 remaining seats.
So, we can seat the three girls is (9)(8)(7) ways

At this point, throw away the remaining empty seats, and you have 11 children seated.

So, the total number of ways to seat all of the boys and girls is (8!)(9)(8)(7)

Almost done....

P(no girls together) = [# of ways to seat everyone with no girls together]/[# of ways to seat everyone]
= (8!)(9)(8)(7)/11!
= 28/55

Cheers,
Brent