Question

Binary Codes Using the digits $0$and $1$, how many different numbers consisting of $8$digits can be formed?

Short answer

The answer is$256.$

Step-by-step solution

Step 1. Given information.

Digits $0$and $1$are given.

Step 2. Calculation.

We are choosing $8$digit number by$0$and$1.$

$8$digits are chosen from $2$distinct objects.

Thus use the method of distinct object with repetition is $n^r.$

So,

$$\begin{gathered} n^r=2^8 \\ =2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ =256 \end{gathered}$$