Originally Posted by

**galactus** I would like to extend our old circular permutation problem to 10 boys and 4 girls.

That is, "how many arrangements are possible with 10 boys and 4 girls if no 2 girls can set together?".

Anyone wanna tackle this one. It's a little rougher than the 5 boys and 3 girls.

We know there are 13! possible arrangements, altogether. Big number.

Extending the concept of counting orbits you have,

$\displaystyle \frac{1}{14}S\cdot 4!\cdot 10!$

Where, $\displaystyle S$

The the number of different position without rotations, for example,

Code:

G*G*G*G******
G*G*G**G*****
G**G**G**G***
AND SO ON

The problem reduces to finding S.