parveen110 wrote: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
