An engineer is required to visit a factory for exactly four days...

Selection of $4$ non-consecutive days out of $15$ days is same as first remove $4$ objects from $15$ objects, so remainging objects are $11$ having $12$ gaps in which these $4$ objects can be placed again.

So, total number of ways is $C4 12 = \dfrac{12 . 11 . 10 . 9}{4 . 3 . 2 . 1} = 495$.