Given N bits, how many integers can be represented in binary?

Question

Suppose you have 14 bits. How do you determine how many integers can be represented in binary from those 14 bits?

Is it simply just 2^n? So 2^14 = 16384?

Please

Answer 1

The answer depends on whether you need signed or unsigned integers.

If you need unsigned integers then using 2^n you can represent integers from 0 to 2^n exclusive. e.g. n=2; 2^2=4 you can represent the integers from 0 to 4 exclusive (0 to 3 inclusive). Therefore with n bits, you can represent a maximum unsigned integer value of 2^n - 1, but a total count of 2^n different integers including 0.

If you need signed integers, then half of the values are negative and half of the values are positive and 1 bit is used to indicate whether the integer is positive or negative. You then calculate using using 2^n/2. e.g. n=2; 2^2/2=2 you can represent the integers from -2 to 2 exclusive (-2 to +1 inclusive). 0 is considered postive, so you get 2 negative values (-2, -1) and 2 positive values (0 and +1). Therefore with n bits, you can represent signed integers values between (-) 2^n/2 and (+) 2^n/n - 1, but you still have a total count of 2^n different integers as you did with the unsigned integers.

Answer 2

Yes, it's that easy as 2^n.

A bit can have 2 distinct values: 0 and 1.

If you have 2 bits, than you have 4 distinct values: 00, 01, 10, 11. The list goes on.

Combinatorics has the simple counting formula

N = n_1 ⋅ n_2 ⋅ ... ⋅ n_k

Since n_1 = n_2 = n_k = 2 you can reduce the formula to

N = 2 ^ k