In how many ways can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots?

Correct Answer: (B) 19

Let $x, y, z$$\ge$ 0 denote the number of singles, fours, and sixes scored. The condition is $x + 4y + 6z = 25$, which is equivalent to finding the number of non-negative integer pairs $(y, z)$ satisfying $4y + 6z$$\le$ 25. Testing values of $z$: For $z = 0$, $4y$$\le$ 25 $\implies$ y $\in$ \{0, 1, 2, 3, 4, 5, 6\} (7 values). For $z = 1$, $4y$$\le$ 19 $\implies$ y $\in$ \{0, 1, 2, 3, 4\} (5 values). For $z = 2$, $4y$$\le$ 13 $\implies$ y $\in$ \{0, 1, 2, 3\} (4 values). For $z = 3$, $4y$$\le$ 7 $\implies$ y $\in$ \{0, 1\} (2 values). For $z = 4$, $4y$$\le$ 1 $\implies$ y = 0 (1 value). Total ways $= 7 + 5 + 4 + 2 + 1 = 19$.