Let the singles taken and the fours and the sixes scored by the batsman be a, b and c, respectively. So, as per the question:

a + 4b + 6c = 25, wherein a, b, c ≥ 0

If no six has been hit, i.e. c = 0 a + 4b = 25
So, the possible values of (a, b) may be (1, 6), (5, 5), (9, 4), (13, 3), (17, 2), (21, 1), (25, 0), i.e. 7 possible ways.

If one six has been hit, i.e. c = 1 a + 4b = 19
So, the possible values of (a, b) may be (3, 4), (7, 3), (11, 2), (15, 1), (19, 0), i.e. 5 possible ways.

If two sixes have been hit, i.e. c = 2a + 4b = 13
So, the possible values of (a, b) may be (1, 3), (5, 2), (9, 1), (13, 0), i.e. 4 possible ways.

If three sixes have been hit, i.e. c = 3a + 4b = 7
So, the possible values of (a, b) may be (3, 1), (7, 0), i.e. 2 possible ways.

If four sixes have been hit, i.e. c = 4a + 4b = 1
So, the possible values of (a, b) may be (1, 0), i.e. 1 possible way.
So, total number of possible ways = 7 + 5 + 4 + 2 + 1 = 19