The digits 1-5 are used for a set of locker codes. suppose the digits cannot repeat. find the number of possible two digit and three digit codes. describe any pattern and use it to predict the number of possible five digit codes.

Two-digit codes:

The first digit can be any one of 5. For each of those . . .
the second digit can be any one of the remaining 4.

Total 2-digit possibilities  =  (5 x 4) = 20 .

Three-digit codes:
The first 2 digits are as above.
The 3rd digit can be any one of the remaining 3.

Total 3-digit possibilities  =  (5 x 4 x 3) = 60 .

The pattern is:

    The number of n-digit possibilities =    5 ! / (5 - n) !

(The symbol " ! " means "factorial" ... the product of all
the whole numbers from 1 up to the number.)

Five-digit codes:      Using the pattern ...

Number of 5-digit possibilities =

           (5 !) / (5 - 5) !  =  120 / 0 !

That ' 0 ! ' is awkward, and hard to explain.  For the time being,
I'll just have to ask you to accept that ' 0 ! ' is defined as ' 1 ' .
Then the number of 5-digit possibilities is  120 .