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?

Let x be the number of singles, y be the number of fours, and z be the number of sixes. The problem requires finding the number of non-negative integer solutions to the equation: 1x + 4y + 6z = 25 We can systematically find the solutions by iterating through the possible values of z (number of sixes), as it has the largest coefficient and thus the fewest possibilities. Case 1: z = 0 (No sixes) The equation becomes x + 4y = 25. Possible values for y (number of fours): If y = 0, x = 25 (25 singles) If y = 1, x = 21 (21 singles, 1 four) If y = 2, x = 17 (17 singles, 2 fours) If y = 3, x = 13 (13 singles, 3 fours) If y = 4, x = 9 (9 singles, 4 fours) If y = 5, x = 5 (5 singles, 5 fours) If y = 6, x = 1 (1 single, 6 fours) (If y = 7, 4y = 28 > 25, so not possible) This gives 7 ways. Case 2: z = 1 (One six) The equation becomes x + 4y + 6(1) = 25, which simplifies to x + 4y = 19. Possible values for y: If y = 0, x = 19 (19 singles, 1 six) If y = 1, x = 15 (15 singles, 1 four, 1 six) If y = 2, x = 11 (11 singles, 2 fours, 1 six) If y = 3, x = 7 (7 singles, 3 fours, 1 six) If y = 4, x = 3 (3 singles, 4 fours, 1 six) (If y = 5, 4y = 20 > 19, so not possible) This gives 5 ways. Case 3: z = 2 (Two sixes) The equation becomes x + 4y + 6(2) = 25, which simplifies to x + 4y = 13. Possible values for y: If y = 0, x = 13 (13 singles, 2 sixes) If y = 1, x = 9 (9 singles, 1 four, 2 sixes) If y = 2, x = 5 (5 singles, 2 fours, 2 sixes) If y = 3, x = 1 (1 single, 3 fours, 2 sixes) (If y = 4, 4y = 16 > 13, so not possible) This gives 4 ways. Case 4: z = 3 (Three sixes) The equation becomes x + 4y + 6(3) = 25, which simplifies to x + 4y = 7. Possible values for y: If y = 0, x = 7 (7 singles, 3 sixes) If y = 1, x = 3 (3 singles, 1 four, 3 sixes) (If y = 2, 4y = 8 > 7, so not possible) This gives 2 ways. Case 5: z = 4 (Four sixes) The equation becomes x + 4y + 6(4) = 25, which simplifies to x + 4y = 1. Possible values for y: If y = 0, x = 1 (1 single, 4 sixes) (If y = 1, 4y = 4 > 1, so not possible) This gives 1 way. Case 6: z = 5 (Five sixes) 6(5) = 30, which is greater than 25. So, no solutions are possible for z >= 5. Total number of ways = Sum of ways from all cases = 7 + 5 + 4 + 2 + 1 = 19 ways. The final answer is B