Let's think about this another way. We can't have a zero and we can't have digits repeated. So all these numbers will come from the set {1, 2, 3, ..., 8, 9}.
We need to choose 3 digits from there and then arrange them in ascending order.
This is 9 choose 3 = 9! / 3! (9-3)!
= 84 rising numbers
Let's see if we can list them...
123, 124, 125, 126, 127, 128, 129, 134, 135, 136, 137, 138,
139, 145, 146, 147, 148, 149, 156, 157, 158, 159, 167, 168,
169, 178, 179, 189, 234, 235, 236, 237, 238, 239, 245, 246,
247, 248, 249, 256, 257, 258, 259, 267, 268, 269, 278, 279,
289, 345, 346, 347, 348, 349, 356, 357, 358, 359, 367, 368,
369, 378, 379, 389, 456, 457, 458, 459, 467, 468, 469, 478,
479, 489, 567, 568, 569, 578, 579, 589, 678, 679, 689, 789
Indeed 84 numbers!
2007-10-17 13:22:40
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answer #1
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answered by Puzzling 7
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