Answer:
Mason has
- 16 20¢ stamps;
- 25 3¢ stamps;
- 40 49¢ stamps.
Explanation:
Let x be the number of 20¢ stamps, then the number of 3¢ stamps is x + 9.
Mason has 56 total 49¢ and 20¢ stamps, then the number of 49¢ stamps is 56 - x.
Amounts:
- in 49¢:
cents; - in 20¢:
cents; - in 3¢:
cents.
Total amount:

- $23.55.
Hence,

Mason has
- 16 20¢ stamps;
- 25 3¢ stamps;
- 40 49¢ stamps.