How many 3 - digit even number can you form such that if one of the digits is 5, the following digit must be 7?

For a number to be even, its unit digit should be 0,2,4,6,8
Case 1: One of the digit is 5
Hence according to question, 5 can't come in middle and at unit's place, so numbers will be 570,572,574,576,578.
Case 2: No digit is 5
Hence the hundreds place can be filled in 8 ways (except 0,5) and tens place can be filled in 9 ways (except 5).
Number of ways = 8 * 9 * 5 = 360
Hence total number of ways = 360 + 5 = 365