The number of ways, in which 6 boys and 5 girls can sit at a round table, if no two girls are to sit together, is

With 6 distinguishable boys, the number of distinct circular arrangements is $(6-1)! = 5! = 120$.

These boys create 6 distinct spaces for seating the 5 distinguishable girls, ensuring no two girls are adjacent. The number of ways to choose 5 of the 6 spaces is $C(6,5) = 6$. The number of ways to arrange the 5 girls among the chosen spaces is $5! = 120$.

The total number of seating arrangements is the product: $5! \times C(6,5) \times 5! = 120 \times 6 \times 120 = 86400$.