In the binary number system which is used in computer operations, there are only two digits allowed: $0$ and $1$ . If eight binary digits are used, how many different binary numbers can be formed?

Hint: First, we shall analyze the given data so that we can able to solve the problem. It is given that there are only two digits $0$ and $1$are allowed in the binary number system. We need to calculate the various binary numbers that can be formed when we use eight binary digits. We need to start with one-digit, and we shall proceed to get the required answer.

Complete step-by-step answer:
We all know that in the binary number system $0$ and $1$ are the only two digits that are allowed in computer operations.
If we are using eight binary digits, then we need to find how many binary numbers may be formed.
First, let us consider we are using a one-digit number.
Then binary numbers $0$ and $1$ may be formed.
That is the number of possibilities when we have a $1$ -digit number is $2$
Now, let us consider that we are using a two-digit number.
Then the binary numbers $01,01,10$ and $11$ may be formed.
Since we have a two-digit number, the number of possibilities will be $2 \times 2 = {2^2} = 4$
Now, we shall consider that we are using a three-digit number.
Then the binary numbers $000,001,010,100,011,101,110,111$ will be formed.
That is number of binary numbers formed is $8$
We can also say that we have a three-digit number, and then the number of possibilities will be ${2^3} = 8$
We shall note that when we increase the digits, the power upon $2$ is increasing according to the digits.
We can analyze that the power of $2$ and the number of digits are the same.
That is, if we have a one-digit number, then we may get ${2^1}$ binary numbers.
If we have a two-digit number, then we may get ${2^2}$ binary numbers.
According to the question, if we have an eight-digit number, the number of possibilities may be ${2^8} = 256$ binary numbers.
Therefore, if eight binary digits are used, then $256$ binary numbers can be formed.

Note: We can note that there is a general form for calculating the various binary numbers that can be formed.
That is, ${2^n}$ binary numbers can be formed, where $n$ is the number of binary digits we are using.
Therefore, if eight binary digits are used, ${2^8} = 256$ binary numbers can be formed.