An expert question on combinations for P5–P6, with a step-by-step solution, diagrams and the answer. Counting arrangements and choices appears in the Hua Xia Cup from the P1 and P2 first rounds to the P5 finals.
She buys $31 of stationery and wants to pay exactly $31 in coins, with no change; the cashier will take at most 9 coins.
How many different ways can she pay? (Only how many of each coin counts.)
Show the hints (3)
💡Hint 1: Decide the number of ten-dollar coins, then five-dollar coins. The rest is made up in two-dollar and one-dollar coins.
💡Hint 2: Once the ten-dollar and five-dollar coins are set, choosing the number of two-dollar coins fixes the one-dollar coins. Do more two-dollar coins mean more coins in all, or fewer?
💡Hint 3: The number of two-dollar coins must meet three limits at once: at most 3; at most 6 one-dollar coins; at most 9 coins in all. Count case by case.
Worked solution
Have a go first, then read the solution step by step.
1.After t ten-dollar and f five-dollar coins, r dollars remain; with w two-dollar coins the one-dollar coins are r − 2 × w, so the coins number t + f + w + (r − 2 × w) = t + f + r − w. Each extra $2 coin means one coin fewer: at most 9 coins means w is at least t + f + r − 9.
2.2 × $10 and 2 × $5: $1 to go. With at most 6 one-dollar and 3 two-dollar coins, w can be 0; the coin limit changes nothing. So w is 0: 1 way.
3.2 × $10 and 1 × $5: $6 to go. With at most 6 one-dollar and 3 two-dollar coins, w can be 0 to 3; the coin limit changes nothing. So w is 0 to 3: 4 ways.
4.1 × $10 and 3 × $5: $6 to go. With at most 6 one-dollar and 3 two-dollar coins, w can be 0 to 3; the coin limit needs w at least 1 + 3 + 6 − 9 = 1. So w is 1 to 3: 3 ways.
5.In all: 1 + 4 + 3 = 8.
⚠️Common mistake: forgetting the 9-coin limit and answering 11; or getting the direction wrong — w has a lower limit, not an upper one. With too few $10 and $5 coins, the $1 coins push the count past 9.
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