Difference between revisions of "1990 AHSME Problems/Problem 17"

(Solution)
 
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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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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>
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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, $100,101,\cdots,999$ have three different digits in increasing order or in decreasing order?

$\text{(A) } 120\quad \text{(B) } 168\quad \text{(C) } 204\quad \text{(D) } 216\quad \text{(E) } 240$

Solution

For decreasing order, we just need to choose any three digits from the ten: $\tbinom{10}3$

For increasing order, the number cannot start with $0$, so choose from nine: $\tbinom93$

The sum is $204$, giving $\fbox{C}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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