Given,
7 boys and 5 girls are to be seated around a circular such that no two girls to be seated together,
Now we know that n objects can be arranged in a circle in (n−1)! ways.
Let us first arrange 7 boys in circular arrangement in (7−1)! ways.
Now there will be 7 gaps.
So let us select any 5 gaps out of 7 gaps and arrange 5 girls in the chosen gaps. This can be done in C57×5! ways.
Hence, required arrangements are 6!×C57×5!
=6×5!×27×6×5!
=126(5!)2.
Therefore, required arrangements are 126(5!)2