How many of the numbers 100, 101, …, 999 have three different digits in increasing order or in decreasing order?

1990 AMC 12 Problem 17

Attempt Problem 17 of the 1990 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 12 solutions, or check the answer key.

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17.

How many of the numbers

100,

101,

\ldots,

999

have three different digits in increasing order or in decreasing order?

120

168

204

216

240

Choosing three distinct digits fixes their increasing or decreasing order

Treat the digit

0

carefully in the increasing case

An increasing three-digit number cannot use

0,

so there are

\binom93=84

increasing numbers. Any three digits chosen from

0

through

9

form a valid decreasing three-digit number, because the largest digit comes first; this gives

\binom{10}{3}=120.

The total is

84+120=204.

Thus the correct answer is C.