Binary Counting : How many values can be represented w/ 9 bits?

Question

I fear that I may be over thinking this homework question that I have for my assembly course. It reads:

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?

Answer 1

What you're missing: Zero is a value

Answer 2

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)?

Answer 3

2⁹ = 512 values, because that's how many combinations of zeroes and ones you can have.

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What those values represent however will depend on the system you are using. If it's an unsigned integer, you will have:

000000000 = 0 (min)
000000001 = 1
...
111111110 = 510
111111111 = 511 (max)

In two's complement, which is commonly used to represent integers in binary, you'll have:

000000000 = 0
000000001 = 1
...
011111110 = 254
011111111 = 255 (max)
100000000 = -256 (min) <- yay integer overflow
100000001 = -255
...
111111110 = -2
111111111 = -1

In general, with k bits you can represent 2^k values. Their range will depend on the system you are using:

Unsigned: 0 to 2^k-1
Signed: -2^k-1 to 2^k-1-1

Answer 4

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

Answer 5

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)

Answer 6

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?

Answer 7

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.