Suppose that a code consists of 5 digits, none of which is repeated. (A digit is one of the 10 numbers 0,1,2,3,4,5,6,7,8,9.) How many codes are possible?

30240

Step-by-step explanation:

There are two primary ways to approach this problem - the logical and the mathematical.

The logical way is very simple:

Since you know that there are 5 digits and each must be different, start with the first digit and work your way up.

The first digit can be any of 10 digits (0-9).

However, the second digit can be any of only 9 digits as 1 digit has already been used and cannot be repeated.

Similarly, applying the same logic, there are only 8 options for the third digit, 7 for the fourth digit, and 6 for the fifth and final digit.

Now, multiply all these numbers together to get the total number of configurations like so:

10⋅9⋅8⋅7⋅6=30240

The more mathematically inclined way requires the use of a special formula. Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:

P (n, r) = n! (n-r) !

where n is the number of numbers in the set and r

is the subset.

Since there are 10 digits to choose from, we can assume that n=10

.

Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r=5

.

Now, plug these values into the formula and solve:

=10! (10-5) !

=10!5!

=10⋅9⋅8⋅7⋅6

=30240