A code on a student ID card begins with 3 letters and ends with 5 digits. How many different ID codes are possible if ... the letters and digits cannot be repeated?

The total number of combinations is 2,358,720,000

Step-by-step explanation:

For the letter part of the ID we have 3 letters out of a space of 26 possible letters (a-z), and they can't repeat. For the number part we want to group 5 numbers out of 10 possible algarisms (0-9). So we can make an arrangement for the letters and one for the numbers and multiply them. The arrangment can be done using the following formula:

A (n, k) = (n!) / (n-k) !

Where n is the total number of possibilities and k is the size of the group.

For the letters:

A (26,3) = (26!) / (26-3) ! = (26!) / (23!) = (26*25*24*23!) / (23!) = 26*25*24 = 15600

For the numbers:

A (10,5) = (10!) / (10 - 5) ! = 10!/5! = (10*9*8*7*6*5!) / (5!) = 10*9*8*7*6*5 = 151200

The total number of combinations is the product of both, so:

combinations = 15600*151200 = 2,358,720,000