How many even numbers greater than 300 can be formed with

Not a GMAT question.

Number can be 3, 4, or 5 digit (4 and 5 digit number will obviously be more than 300).

5 digit even numbers:

Last digit must be either 2 or 4 (for number to be even), so 2 choices. First 4 digits can be arranged in 4! # of ways. So total # of 5 digit even numbers possible is \(4!*2=48\);

4 digit even numbers:

Last digit again must be either 2 or 4, so 2 choices. # of ways to choose rest 3 digits out of 4 digits left when order of the digits matters is \(P^3_4\). So total # of 4 digit even numbers possible is \(P^3_4*2=48\);

3 digit even numbers:

Last digit again must be either 2 or 4, so 2 choices. # of ways to choose rest 2 digits out of 4 digits left when order of the digits matters is \(P^2_4\). So total # of 3 digit even numbers possible is \(P^2_4*2=24\).

But out of these 24 3-digit even numbers some will be less than 300, the ones starting with 1 or 2. There are 9 such numbers (1X2 and 3 options for the second digit X, so 3 plus 1X4 and 3 options for X, so 3 plus 2X4 and 3 options for X, so 3), which means that # of 3 digit even numbers more than 300 is \(24-9=15\);

So total \(48+48+15=111\).