How many ways are there to arrange 6 girls and 15 boys in a circle such that there are at least 2 boys between any 2 adjacent girls? Please help me how to proceed with this.
Arrange first the girls in $5!$ ways, now we must choose $12$ boys from the $15$ in $15 \choose 12$ ways, and these boys will be arranged such that each girl has $2$ boys next to them in $12!$ Ways, then the remaining $3$ boys are left. We have already placed $18$ people, so we have $19$ position left for first boy, then $20$ position for 2nd boy and $21$ position for 3rd. So I get it as,
$5!$$15 \choose 12$($12!$)($19*20*21$)
But this looks clumsy and I'm not sure if its correct.
I searched for similar circular permutation questions on this site, but it was mostly regarding equal number of boys and girls, and since this question is different I couldn't co relate. If question is repeated, apologies and kindly redirect me to such duplicate question, else please help me find the solution. Thank you