Problem 12 · 1990 AJHSME
Hard
Counting & Probability
ordering-permutations
There are twenty-four 4-digit numbers that use each of the four digits 2, 4, 5, and 7 exactly once. Listed in numerical order from smallest to largest, the number in the 17th position in the list is
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Answer: B — 5724.
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Hint 1 of 2
Listing all 24 numbers in order is slow and error-prone. Instead think in *blocks*: once you fix the first digit, the other three digits can be arranged 6 ways (3×2×1). So the smallest 6 numbers all start with 2, the next 6 start with 4, and so on.
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Hint 2 of 2
Walk the blocks of 6 to land on the right leading digit, then list only inside that small block. This is counting-by-blocks (place-value with permutations).
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Approach: count in blocks of 6, then order only inside the right block
- Fixing the first digit leaves 3 digits to arrange = 3×2×1 = 6 numbers per block, in increasing leading digit (2, 4, 5, 7). So: positions 1–6 start with 2, 7–12 start with 4, 13–18 start with 5, 19–24 start with 7.
- The 17th falls in the '5' block (13–18), and it's the 17−12 = 5th number in that block.
- List just the 5-block in order (remaining digits 2,4,7): 5247, 5274, 5427, 5472, 5724, 5742. The 5th is 5724.
- *Why this transfers:* to find the k-th item in an ordered list of arrangements, never write them all — divide into blocks (here 6 per leading digit), jump to the right block, and only sort the few inside it.
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