Question

Assume you have a safe code that is 8 digits long. The numbers 0-9 may be used for any digit. How many possible codes are there? Show your work and leave your answer in either exponential or factorial form.

Answer 1

Answer:

10^8 choices

Step-by-step explanation:

You have 10 numbers to choose from for each digit of the code.

There are 8 digits of the code.

You have 10*10*10*10*10*10*10*10 = 10^8 choices in total.

Answer 2

Number of possible codes are  $10^{8}$

What is Permutation?

In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements

Given,

Number of digits in passcode = 8

The numbers 0-9 may be used for any digit. Therefore total of 10 numbers

Here repetition is allowed.

Therefore, each of the 8 vacant place can be filled in the succession of 10 different ways

Number of possible codes are = 10×10×10×10×10×10×10×10 = $10^{8}$

Hence, the number of possible codes are  $10^{8}$

Learn more about Permutation here

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