My Approach:
Restrictions:
1. First digit should not be 0 (9 possibilities: 1,2,3,4,5,6,7,8,9)
2. Last digit should not be odd (5 possibilities 0,2,4,6,8)
Lets take 2 Cases, when 0 is the last digit and when 0 is not the last digit
Case 1: When 0 is the last digit
4th digit - 0 (1 possibility)
1st digit - Any number 1-9 (9 possibilities)
2nd digit - Any remaining number out of 10 after 2 numbers are selected (8 possibilities)
3rd digit - Any remaining number out of 10 after 3 numbers are selected (7 possibilities)
Hence possible cases become
9 *
8 *
7 *
1 = 504
NOTE: First fill the possibilities where restrictions are applied
Case 2: When 0 is not the last digit
4th digit - 2,4,6,8 (4 possibilities)
1st digit - Any number from 1-9 except 2,4,6,8 (8 possibilities since only one of 2,4,6,8 will be used at a time)
2nd digit - Any remaining number out of 10 after 2 numbers are selected (8 possibilities)
3rd digit - Any remaining number out of 10 after 3 numbers are selected (7 possibilities)
Hence possible cases become
8 *
8 *
7 *
4 = 1792
Total becomes = 1792 + 504
Since number of students are 500 less, hence number of students = 1792 + 504 - 500 = 1796
Hope it helps!