Choose from 1,2,4,5.

We can have 1 digit, 2 digit, 3 digit and 4 digit numbers. All must be even, > 1.

① 1 digit

Only 2 and 4
Number of ways = 2


② 2 digits

Last digit is 2 or 4. That's 2 ways. So just need to pick 1 out of the 3 remaining digits
for the tens place.

Number of ways = 2 x 3C1
= 2 x 3
= 6

③ 3 digits

Last digit is 2 or 4. That's 2 ways. Just need to pick 2 out of 3 remaining digits for the tens and hundreds place. Then permuate these 2 since we can switch them around.


Number of ways = 2 x 3C2 x 2!
= 2 x 3 x 2
= 12


③ 4 digits

Last digit is 2 or 4. That's 2 ways. The 3 remaining digits go into the tens, hundreds and thousands place.

We can switch them around. Just need to permutate these 3.

Number of ways = 2 x 3!
= 2 x 6
= 12

Total = 12 + 12 + 6 + 2 = 32

The idea goes like this. This question is made easy since there is no repetition of digits.

Even numbers must end in 2 or 4 considering that the list contains only 1, 2, 4 and 5.

****Case 1: One-digit numbers****

A no brainer situation, there are 2 possible one-digit numbers.

****Case 2: Two-digit numbers****

The ones digit must be 2 or 4 (two choices). Either way, the tens digit can contain any of the three unused numbers (three choices). There are 2 x 3 = 6 possible two-digit numbers.

****Case 3: Three-digit numbers****

The ones digit must be 2 or 4 (two choices). Either way, the tens digit can be occupied by any of the three unused numbers (three choices) and thereafter, the hundreds digit can be occupied by any of the two remaining unused digits (two choices).

There are 2 x 3 x 2 = 12 possible three-digit numbers.

****Case 4: Four-digit numbers****

The ones digit must be 2 or 4 (two choices). Either way, the tens digit can be occupied by any of the three unused numbers (three choices) and thereafter, the hundreds digit can be occupied by any of the two remaining unused digits (two choices) and finally, the thousands digit is left with only one choice.

There are 2 x 3 x 2 x 1 = 12 possible four-digit numbers.

****Overall cases****

Total possible number of cases is 2 + 6 + 12 + 12 = 32 different even numbers.

These are the possible numbers if you list them out stubbornly.

(1-digit) 2, 4

(2-digit) 12, 14, 24, 42, 52, 54

(3-digit) 124, 142, 152, 154, 214, 254, 412, 452, 512, 514, 524, 542

(4-digit) 1254, 1452, 1524, 1542, 2154, 2514, 4152, 4512, 5124, 5142, 5214, 5412

For a total of 32 possible numbers. This method is not recommended because it gets exponentially long-winded with more digits.

Brute force method won't get the full marks