In a group of 21 persons, every two person communicate with different radio frequency. It's possible for two person to not communicate (means there's no frequency occupied to connect them). Only one frequency used by each couple, and it's unique for every couple. In every 3 persons, exactly two of them is not communicating to each other. Determine the maximum number of frequency required for this group. Explain your answer.
1. **Model the Problem as a Graph:**
   - Let each of the 21 persons be represented by a vertex in a graph.
   - An edge between two vertices indicates that the two corresponding persons communicate with each other using a unique frequency.

2. **Graph Properties:**
   - Given that in every subset of 3 persons, exactly two of them do not communicate with each other, we can infer that the graph is a complete bipartite graph \( K_{m,n} \) where \( m + n = 21 \).

3. **Bipartite Graph Explanation:**
   - In a bipartite graph \( K_{m,n} \), vertices are divided into two disjoint sets such that no two vertices within the same set are adjacent.
   - This ensures that in any subset of 3 vertices, there will be at least one pair of vertices that are not connected, satisfying the given condition.

4. **Determine the Maximum Number of Frequencies:**
   - The maximum number of edges (frequencies) in a bipartite graph \( K_{m,n} \) is given by \( m \times n \).
   - To maximize \( m \times n \) under the constraint \( m + n = 21 \), we need to find the values of \( m \) and \( n \) that maximize the product.

5. **Maximizing the Product \( m \times n \):**
   - Let \( m = 11 \) and \( n = 10 \) (or vice versa, since the product is symmetric).
   - The product \( 11 \times 10 = 110 \).

6. **Conclusion:**
   - The maximum number of unique frequencies required is 110.

The final answer is \( \boxed{110} \).