Difference between revisions of "2006 AMC 12B Problems/Problem 9"
(alternate solution) |
m (→Solution 2) |
||
| Line 19: | Line 19: | ||
===Solution 2=== | ===Solution 2=== | ||
| − | The last digit is | + | The last digit is 4, 6, or 8. |
If the last digit is <math>x</math>, the possibilities for the first two digits correspond to 2-element subsets of <math>\{1,2,\dots,x-1\}</math>. | If the last digit is <math>x</math>, the possibilities for the first two digits correspond to 2-element subsets of <math>\{1,2,\dots,x-1\}</math>. | ||
| − | Thus the answer is <math> | + | Thus the answer is <math>{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = 34</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | ||
Revision as of 18:31, 5 January 2009
Problem
How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?
Solution
Solution 1
Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 8, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
and so on.
So, the answer is 
.
Solution 2
The last digit is 4, 6, or 8.
If the last digit is 
, the possibilities for the first two digits correspond to 2-element subsets of 
.
Thus the answer is 
.
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
| All AMC 12 Problems and Solutions | |
