Circular permutation on seating There are 21 people ,15 boys and 6 girls.how many ways are there to seat at least 2 boys between any two adjacent girls in a round table?. 
I get my answer 708480.i m wrong ,i think.please help me.

Distinguishing only by gender, with the usual assumption of unnumbered seating since it is not explicitly specified otherwise, the problem simplifies to placing $15$ green marbles and $6$ red marbles on an unnumbered circle with the given stipulations.
The $6$ red marbles in a circle will have $6$ gaps or "compartments", in which we can initially place $2$ green marbles each, with the remaining $3$ green marbles placements just enumerated to be $10$:
$|\; 3\; |\; 0\; |\; 0\; |\; 0\; |\; 0\; |\; 0 \; |\; (3)\quad\quad|\; 2\; |\; 1\; |\; 0\; |\; 0\; |\; 0\; |\; 0 \; |\; (2)\quad\quad|\; 2\; |\; 0\; |\; 1\; |\; 0\; |\; 0\; |\; 0 \; |\; (2)$
$|\; 2\; |\; 0\; |\; 0\; |\; 1\; |\; 0\; |\; 0 \; |\; (2)\quad\quad|\; 2\; |\; 0\; |\; 0\; |\; 0\; |\; 1\; |\; 0 \; |\; (2)\quad\quad|\; 2\; |\; 0\; |\; 0\; |\; 0\; |\; 0\; |\; 1 \; |\; (2)$
$|\; 1\; |\; 1\; |\; 1\; |\; 0\; |\; 0\; |\; 0 \; |\; (1)\quad\quad|\; 1\; |\; 1\; |\; 0\; |\; 1\; |\; 0\; |\; 0 \; |\; (1)\quad\quad|\; 1\; |\; 1\; |\; 0\; |\; 0\; |\; 1\; |\; 0 \; |\; (1)$
and the last one,$\;|\; 1\; |\; 0\; |\; 1\; |\; 0\; |\; 1\; |\; 0 \; |\; (1)$
