- A. 18
- B. 19
- C. 20
- D. 21
Answer: B
Explanation
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 x + 4y + 6z = 25. We can systematically find combinations by iterating through possible values of z (number of sixes):
1. If z = 0: x + 4y = 25. Possible (x, y) pairs are (1, 6), (5, 5), (9, 4), (13, 3), (17, 2), (21, 1), (25, 0). (7 ways)
2. If z = 1: x + 4y = 19. Possible (x, y) pairs are (3, 4), (7, 3), (11, 2), (15, 1), (19, 0). (5 ways)
3. If z = 2: x + 4y = 13. Possible (x, y) pairs are (1, 3), (5, 2), (9, 1), (13, 0). (4 ways)
4. If z = 3: x + 4y = 7. Possible (x, y) pairs are (3, 1), (7, 0). (2 ways)
5. If z = 4: x + 4y = 1. Possible (x, y) pair is (1, 0). (1 way)
For z ≥ 5, 6z would be 30 or more, exceeding 25, so no solutions exist. Summing the ways from each case: 7 + 5 + 4 + 2 + 1 = 19 ways. This question tests basic combinatorics and systematic enumeration, a common type in CSAT.