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.
The question
Concert tickets are numbered 1 to 3004, one ticket per number.
A ticket whose number contains each of the digits 1, 2, 3 exactly once wins a souvenir.
For example 312 and 1023 do, but 121 doesn't (1 appears twice and there's no 3).
How many tickets win a souvenir?
Show the hints (3)
💡Hint 1: Sort by number of digits. Can a one- or two-digit number work? How many three-digit ones are there?
💡Hint 2: Sort the four-digit ones by the thousands digit: one of 1, 2, 3, or not — they count differently. Can the fourth digit be 0?
💡Hint 3: Below 3 in the thousands place, the last three digits can be anything; with 3 in the thousands place the number can't pass 3004, so the hundreds digit must be 0.
Worked solution
Have a go first, then read the solution step by step.
1.One- and two-digit numbers can't hold three different digits. Three-digit: 1, 2, 3 in some order, 3 × 2 × 1 = 6.
2.Four-digit with thousands digit 1, 2 (one of 1, 2, 3): the other two and one more digit fill the last three places. The extra digit isn't 1, 2, 3 but may be 0: 7 choices; it goes in 3 places and the other two in 2 orders: 3 × 2 × 7 = 42. In all 2 × 42 = 84.
3.No thousands digit below 3 is outside 1, 2, 3.
4.Thousands digit 3: up to 3004 means a 0 in the hundreds place, so the other two of 1, 2, 3 take the tens and ones: 3012, 3021, of which 0 are at most 3004.
5.In all: 6 + 84 + 0 + 0 = 90.
⚠️Common mistake: not allowing the extra digit to be 0, counting only 3 × 2 × 6 = 36 for each thousands digit in 1, 2, 3 and getting 78; or ignoring the bound 3004 for thousands digit 3.
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