Step 1: Understanding the Question:
Find circular permutations where a specific group of 3 people is not kept as a contiguous block.
Step 2: Key Formula or Approach:
1. Total circular arrangements of $n$ people $= (n-1)!$.
2. Arrangements with specified group together $= (\text{Total items} - \text{group size})! \times (\text{group size})!$.
Step 3: Detailed Explanation:
Total people $= 6 + 4 = 10$.
Total circular arrangements $= (10 - 1)! = 9! = 362880$.
Let the "special" people be $B_1, B_2, G_1$. To find the number of ways they sit together, treat them as one unit.
Remaining entities $= (10 - 3) = 7$ plus the unit $= 8$ entities.
Arrangements of 8 entities in a circle $= (8 - 1)! = 7!$.
Internal arrangements of the special 3 people $= 3! = 6$.
Total ways they sit together $= 7! \times 6 = 5040 \times 6 = 30240$.
Ways they never sit together $= \text{Total} - \text{Ways they sit together}$
\[ 362880 - 30240 = 332640 \]
Step 4: Final Answer:
The number of ways is $332640$.