Difference between revisions of "2002 AMC 10P Problems/Problem 3"
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Then the second <math>1</math> has <math>4</math> places. | Then the second <math>1</math> has <math>4</math> places. | ||
| − | <math> | + | <math> \dots </math> |
This continues as the answer comes to | This continues as the answer comes to | ||
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<math> 5 + 4 + 3 + 2 + 1 = 15.</math> | <math> 5 + 4 + 3 + 2 + 1 = 15.</math> | ||
| − | Thus, our answer is < | + | Thus, our answer is <math>\boxed{\textbf{(D) } 15}.</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=2|num-a=4}} | {{AMC10 box|year=2002|ab=P|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:45, 14 July 2024
Problem
Mary typed a six-digit number, but the two 
s she typed didn't show. What appeared was 
How many different six-digit numbers could she have typed?
Solution 1
We can split this into a little bit of casework which is easy to do in our head.
Case 1: The first is ahead of the first 
Then the second has 
places.
Case 2: The first is below the first
Then the second has 
places.
This continues as the answer comes to
Thus, our answer is 
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
