Let the singles taken and the fours and the sixes scored by the batsman be x, y and z respectively.

So, as per the question: $x + 4y + 6z = 25$, wherein $x, y, z \ge 0$

If no six has been hit, i.e. $z = 0$ $x + 4y = 25$

So, the possible values of (x, y) 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. $z = 1$ $x + 4y = 19$

So, the possible values of (x, y) 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. $z = 2$ $x + 4y = 13$

So, the possible values of (x, y) may be (1, 3), (5, 2), (9, 1), (13, 0), i.e. 4 possible ways.

If three sixes have been hit, i.e. $z = 3$ $x + 4y = 7$

So, the possible values of (x, y) may be (3, 1), (7, 0), i.e. 2 possible ways.

If four sixes have been hit, i.e. $z = 4$ $x + 4y = 1$

So, the possible values of (x, y) may be (1, 0), i.e. 1 possible way.

So, total number of possible ways = $7 + 5 +4 + 2 + 1 = 19$

Hence, option (B) is correct.