P&C- A row of seats in a movie hall..

[parveen110]

A row of seats in a movie hall contains 10 seats. 3 Girls & 7 boys need to occupy those seats. What is the probability that no two girls will sit together?

[Brent@GMATPrepNow]

A row of seats in a movie hall contains 10 seats. 3 Girls & 7 boys need to occupy those seats. What is the probability that no two girls will sit together?

This is a very difficult problem that is probably beyond the scope of the GMAT. What's the source?

That said, here's one 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 10 children, so we can arrange them in 10! ways.

# of ways to seat everyone with no girls together
Take 15 chairs (yes 15), and first seat the 7 boys in chairs 2, 4, 6, 8, 10, 12, and 14
_B_B_B_B_B_B_B_
This can be accomplished 7! ways

Note: This arrangement prevents the girls from sitting together.

Now seat each of the 3 girls in one of the 8 remaining seats.
The first girl can sit in any of the 8 seats.
The second girl can sit in any of the 7 remaining seats.
The third girl can sit in any of the 6 remaining seats.
So, we can seat the three girls in (8)(7)(6) ways

At this point, throw away the remaining empty seats, and you have 10 children seated.

So, the total number of ways to seat all of the boys and girls is (7!)(8)(7)(6)

Almost done....

P(no girls together) = [# of ways to seat everyone with no girls together]/[# of ways to seat everyone]
= (7!)(8)(7)(6)/10!
= [spoiler]7/15[/spoiler]

Cheers,
Brent

[misterholmes]

It's not as bad as all that. Imagine these are the ten seats ##%%%%%%%% where the # mark indicates a girl sitting, and the % indicates an empty seat.
There are 8 arrangements like this one, with the third girl sitting anywhere in the last eight seats. So 8.

Moving the girls down a seat: %##%%%%%%% there are now only 7 uniquely new seating arrangements, since (###%%%%%%%) was counted before.
Now these two hash marks move down the line a total of 8 times, creating 7 uniquely new arrangements each time. So 56 in subtotal.

56 plus the first 8, makes 64 arrangements with at least two girls sitting together.

The total possible arrangements are 10 choose 3. It is easy to see that 64 divided by 10c3 is identical to Brent's answer. Or rather, 1 minus his answer.

Wiki O
4 the GMAT Dojo

[parveen110]

Thanks Brent! I got the question from one of the gmat forums.

misterholmes i din't quite get your meaning. I would like to understand your approach better.
Thanks!!