# Thread: probability of girls sit together/ seperated

1. ## probability of girls sit together/ seperated

Ten Students sit in a row. Find the probability of 6 boys and 4 girls can be arranged if (i) all girls are side by side?
(ii) all girls are seperated by boys

my working for part i is 4!x7!=120960

for part ii, why cant i just take 10! -120960=3507840 ? the ans given is 604800

2. ## Re: probability of girls sit together/ seperated

Hello, delso!

Ten Students sit in a row.
Find the probability of 6 boys and 4 girls can be arranged if

(i) all girls are side by side?

My working: . $7 \times 4!\times 6! \,=\,120,\!960$
Correct!

(ii) all girls are seperated by boys.

Why cant i just take 10! - 120,960=3,507,840? .This is wrong!

The ans given is: 604,800.

You subtracted the number of ways that all four girls are adjacent.

Here is how I handle "not adjacent" problems.

Place the 6 boys in a row.
Insert a space before, after and between them.
. . . $\_\,b\,\_\,b\,\_\,b\,\_\,b\,\_\,b\,\_\,b\,\_$
There are $6!$ permutations of the boys.

Choose 3 of the 7 spaces and assign the girls.
There are $_7P_4$ choices.

Therefore, there are: . $(6!)(_7P_4) \:=\:(720)(840) \:=\:604,\!800$ ways.