Question

Jamie takes part in a quiz competition. she is asked to name the total number of combinations that the bits 1 and 0 will have in a byte. what will be her answer?
A. 256 or 28 different combinations
B. 257 or 28 different combinations
c. 255 or 28 different combinations
D. 255 or 29 different combinations

Answer 1

The answer is A 

I hope this help

Answer 2

Answer: A. $256$ or $2^8$ different combinations

Explanation:

First of all, we have two possible digits or options to choose from, bits 1 and 0. From the question, we know that every combination should form a byte. So, we have to choose eight times (eight bits). For example, a combination could be [ 10001000 ]  another could be [ 00011000 ].

As you can see the order of the digits makes to change the combination. So, byte [ 10000000 ] is different than byte [ 00000001 ]. In the right math language order matter and this is called permutation.  We can say that a permutation is a combinations where order matter.

For a permutation like this, where we can repeat digits, it is possible to calculate the total of permutations by using the formula, $n$ × $n$ × $... (r$ $times) = n^r$

Where:

$n$ is the number of digits or options to choose from

$r$ is the number of digits we choose to make a permutation

From the problem we get that:  

$n=2$ (bits 1 and 0)

$r=8$ (a permutation should form a byte)  

let's replace all values into the formula :

$2$ × $2$ × $... (8$ $times) = 2^8 = 256$  

256 permutations (combinations where order matter)

We can conclude that the total number of combinations that the bits 1 and 0 will have in a byte are A. $256$ or $2^8$ different combinations