The number of ways in which 6 boys and 4 girls can be seated around a round table such that 2 special boys and a special girl never sit together is

Circular arrangements with restrictions are computed using complementary counting. The total arrangements of 10 distinct people around a circular table is $(10-1)! = 9!$.

To count arrangements where two special boys and one special girl always sit together, treat them as a single block. Within this block, they can be arranged in $3!$ ways. With the block considered as one entity, there remain 7 people, giving $(8-1)! = 7!$ circular arrangements. The number of restricted arrangements is therefore $3! \times 7!$.

The number of valid arrangements where the specified individuals never sit together is:

$9! - (3! \times 7!) = 362880 - 30240 = 332640$

The final answer is $\boxed{332640}$.