How many unique possibilities exist in 8 bits (2^8)?

How many unique possibilities exist in 8 bits (2^8)?

• A. 16
• B. 64
• C. 128
• D. 256

Answer: (D)

In an 8-bit binary system, each bit can hold one of two values: 0 or 1. To determine the total number of unique possibilities, you raise the number of possible values (2) to the power of the number of bits (8). This calculation is expressed as 2^8. Performing the calculation: 2^8 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 Thus, the total number of unique combinations that can be represented with 8 bits is 256. This means there are 256 different ways to arrange the bits, ranging from 00000000 (which represents the decimal number 0) to 11111111 (which represents the decimal number 255). Therefore, the choice indicating 256 is the correct answer, accurately reflecting the calculations made based on the properties of binary systems.

In an 8-bit binary system, each bit can hold one of two values: 0 or 1. To determine the total number of unique possibilities, you raise the number of possible values (2) to the power of the number of bits (8). This calculation is expressed as 2^8.

Performing the calculation:

2^8 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

Thus, the total number of unique combinations that can be represented with 8 bits is 256. This means there are 256 different ways to arrange the bits, ranging from 00000000 (which represents the decimal number 0) to 11111111 (which represents the decimal number 255). Therefore, the choice indicating 256 is the correct answer, accurately reflecting the calculations made based on the properties of binary systems.