[binary] How many values can be represented with n bits?

For example, if n=9, then how many different values can be represented in 9 binary digits (bits)?

My thinking is that if I set each of those 9 bits to 1, I will make the highest number possible that those 9 digits are able to represent. Therefore, the highest value is 1 1111 1111 which equals 511 in decimal. I conclude that, therefore, 9 digits of binary can represent 511 different values.

Is my thought process correct? If not, could someone kindly explain what I'm missing? How can I generalize it to n bits?

This question is related to binary bits

The answer is


The thing you are missing is which encoding scheme is being used. There are different ways to encode binary numbers. Look into signed number representations. For 9 bits, the ranges and the amount of numbers that can be represented will differ depending on the system used.


Okay, since it already "leaked": You're missing zero, so the correct answer is 512 (511 is the greatest one, but it's 0 to 511, not 1 to 511).

By the way, an good followup exercise would be to generalize this:

How many different values can be represented in n binary digits (bits)?

Without wanting to give you the answer here is the logic.

You have 2 possible values in each digit. you have 9 of them.

like in base 10 where you have 10 different values by digit say you have 2 of them (which makes from 0 to 99) : 0 to 99 makes 100 numbers. if you do the calcul you have an exponential function

base^numberOfDigits:
10^2 = 100 ;
2^9 = 512

What you're missing: Zero is a value


A better way to solve it is to start small.

Let's start with 1 bit. Which can either be 1 or 0. That's 2 values, or 10 in binary.

Now 2 bits, which can either be 00, 01, 10 or 11 That's 4 values, or 100 in binary... See the pattern?


There's an easier way to think about this. Start with 1 bit. This can obviously represent 2 values (0 or 1). What happens when we add a bit? We can now represent twice as many values: the values we could represent before with a 0 appended and the values we could represent before with a 1 appended.

So the the number of values we can represent with n bits is just 2^n (2 to the power n)