We need to form a group of 3 boys and 3 girls from:

• $10$ boys: $B_1, B_2, \dots, B_{10}$
• $5$ girls: $G_1, G_2, \dots, G_5$

The condition is: $B_1$ and $B_2$ should not be together in the group.

First, count all possible groups without restriction.

Number of ways to choose $3$ boys from $10$: $\binom{10}{3} = 120$

Number of ways to choose $3$ girls from $5$: $\binom{5}{3} = 10$

So total unrestricted groups: $120 \times 10 = 1200$

Now count the groups in which both $B_1$ and $B_2$ are together.

If both $B_1$ and $B_2$ are included among the $3$ boys, then the third boy must be chosen from the remaining $8$ boys: $\binom{8}{1} = 8$

The girls can still be chosen in: $\binom{5}{3} = 10$

Hence, number of invalid groups: $8 \times 10 = 80$

Therefore, required number of groups: $1200 - 80 = 1120$

Final answer: $\boxed{1120}$