Calculating the total number of possibilities in binary?

How would you calculate the tot开发者_运维问答al number of possibilities that binary can have in one byte?

00000000 through 11111111 = num_of_possibilities

The total number is 2 to the power of the number of bits. So, eight bits has 2⁸ possible values.

If you really mean "how to compute it", consider that each bit has two possible values.

So one bit implies 2 values.

Two bits has one set of two values of each possible value of the other bit, so

00
01
10
11

which means a total of 4 (= 2×2) values.

Three bits gives four values twice, or 8 (=4×2) values. Four bits, 8×2; five bits, 16×2, and so on.

So eight bits is 2×2×2×2×2×2×2×2 or 256.

It is a simple question: The number of possibilities is 2ⁿ where n is the number of bits.

So for 1 byte, which is 8 bits, there are 2⁸ possibilites, 256.

There are several methods:

• 2^n where n is the number of bits (2^8) Each bit has 2 possibilities.
• Unsigned value of all 1's + 1 (255 + 1) Count up from 0 to max value (all ones) + zero.
• Build a tree where each leaf is the sum of the values to right and left of the new value from the row above. Possibilities is sum of row having n+1 entries. (2 ( 1 + 8 + 28 + 56 ) + 70) Each value is probability of that number of bits from 0 to n.