Final answer:
Possible combinations of tickets include those where the total number is at least 20 and the total cost doesn't exceed P 12,600. Combinations can range from giving away 28 tickets at P 450.00 each to mixing in more expensive tickets as long as the constraints are met.
Step-by-step explanation:
The student's question involves finding possible combinations of tickets priced at P 450.00 and P 9900.00 that can be given away, given that at least 20 tickets need to be distributed and their total cost should not exceed P 12,600.00. We can approach this problem with a system of inequalities:
Let x be the number of tickets that cost P 450.00, and y be the number of tickets that cost P 9900.00.
- The total number of tickets must be at least 20: x + y ≥ 20.
- The total cost of the tickets must not exceed P 12,600.00: 450x + 9900y ≤ 12,600.
By solving these inequalities, various combinations can be found. For instance, if no expensive tickets are given (y=0), then 28 tickets priced at P 450.00 can be given away (x = 28) because 450 * 28 = 12,600. If at least one expensive ticket is given away (y=1), then only one ticket priced at P 450.00 can be included, satisfying both the total number and cost limitations. Therefore, possible combinations can include (1, 1), (28, 0), (25, 1), (22, 2), etc., ensuring that the constraints are met. These examples use integer values for simplicity, but it should be noted that y will always be less than or equal to 1 due to the high cost of those tickets relative to the total amount allowed.