A processor has 64 general-purpose registers and 50 distinct instruction types. An instruction is encoded in 32-bits. What is the maximum number of bits that can be used to store the immediate operand for the given instruction? ADD R1, #25 // R1 = R1 + 25

To determine the maximum number of bits available for the immediate operand, we find the number of bits required for the other instruction fields:

1. Opcode Field: The processor supports 50 distinct instruction types.

   $$\text{Number of bits for opcode} = \lceil \log_2(50) \rceil = 6\text{ bits}$$

   (since $2^5 = 32 < 50 \le 64 = 2^6$).

2. Register Field: The processor has 64 general-purpose registers.

   $$\text{Number of bits to identify one register} = \log_2(64) = 6\text{ bits}$$

   The instruction ADD R1, #25 specifies only one register operand ($R1$).

3. Immediate Operand Field: The total instruction length is 32 bits. The remaining bits are used for the immediate operand:

   $$\text{Immediate bits} = 32 - (\text{Opcode bits} + \text{Register bits}) = 32 - (6 + 6) = 20\text{ bits}$$

Therefore, the maximum number of bits that can be used to store the immediate operand is 20, which corresponds to option (B).