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

positive

step-by-step explanation:

the quadrant is facing upwards

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x