Difference between revisions of "1990 AHSME Problems/Problem 17"
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | For decreasing order, we just need to choose any three digits from the ten: <math>\tbinom{10}3</math> |
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| + | For increasing order, the number cannot start with <math>0</math>, so choose from nine: <math>\tbinom93</math> | ||
| + | |||
| + | The sum is <math>204</math>, giving <math>\fbox{C}</math> | ||
== See also == | == See also == | ||
Latest revision as of 04:42, 4 February 2016
Problem
How many of the numbers, 
have three different digits in increasing order or in decreasing order?
Solution
For decreasing order, we just need to choose any three digits from the ten: 
For increasing order, the number cannot start with 
, so choose from nine: 
The sum is 
, giving 
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
