Using the digits 1, 2, 3, 4, 5, how many even three-digit numbers less than 500 can be formed if each digit can be used at most once?

Guest Apr 24, 2022

#1**+1 **

There are 4 choices for the left-most digit (1, 2, 3, or 4; 5 can't be chosen)

then there are 4 choices for the middle digit and 3 choices for the right-most digit.

This means that there are 4 x 4 x 3 = 48 choices.

geno3141 Apr 25, 2022

#2**+1 **

The number has to be even, so there are 2 cases: the final digit is a 2, or the final digit is a 4.

If the final digit is a 2:

3 choices for the first number (can't be 2 or 5)

3 choices for the next number (can't be the number just chosen or a 2)

1 choice for the final number (must be a 2)

We can do the same thing for the other case (the final digit is a 4), so there are \(3 \times 3 \times 2 = \color{brown}\boxed{18}\) numbers that work.

BuilderBoi Apr 25, 2022