Difference between revisions of "2002 AMC 10P Problems/Problem 3"
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== Solution 1 == | == Solution 1 == | ||
| + | This is equivalent to <math>{5 \choose 2} + 5</math> since we are choosing <math>5</math> spots (three in between <math>2002</math> and two to the left and right<math> with two </math>1<math>s. We also have to take into account "double </math>1<math>s," namely, </math>112002, 211002, 201102, 200112, and 200211<math> in our </math>5<math> spots. | ||
| + | |||
| + | Thus, our answer is </math>{5 \choose 2} + 5 = \boxed{\textbf{(D) } 15}.<math> | ||
| + | |||
| + | == Solution 2 == | ||
We can split this into a little bit of casework which is easy to do in our head. | We can split this into a little bit of casework which is easy to do in our head. | ||
| − | Case 1: The first <math>1< | + | Case 1: The first </math>1<math> is ahead of the first </math>2.<math> |
| − | Then the second <math>1< | + | Then the second </math>1<math> has </math>5<math> places. |
| − | Case 2: The first <math>1< | + | Case 2: The first </math>1<math> is below the first </math>2.<math> |
| − | Then the second <math>1< | + | Then the second </math>1<math> has </math>4<math> places. |
| − | <math> \dots < | + | </math> \dots <math> |
This continues as the answer comes to | This continues as the answer comes to | ||
| − | <math> 5 + 4 + 3 + 2 + 1 = 15.</math> | + | </math> 5 + 4 + 3 + 2 + 1 = 15.<math> |
| + | |||
| + | Thus, our answer is </math>\boxed{\textbf{(D) } 15}.<math> | ||
| + | |||
| + | == Solution 3 == | ||
| + | We can just count the cases directly since there are so little. | ||
| − | Thus, our answer is <math>\boxed{\textbf{(D) } 15}. | + | Thus, our answer is </math>\boxed{\textbf{(D) } 15}.$ |
== 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 19:23, 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
This is equivalent to 
since we are choosing 
spots (three in between 
and two to the left and right
1
1
112002, 211002, 201102, 200112, and 200211
5$spots.
Thus, our answer is$ (Error compiling LaTeX. Unknown error_msg){5 \choose 2} + 5 = \boxed{\textbf{(D) } 15}.$== Solution 2 == We can split this into a little bit of casework which is easy to do in our head.
Case 1: The first$ (Error compiling LaTeX. Unknown error_msg)1
2.
1
5$places.
Case 2: The first$ (Error compiling LaTeX. Unknown error_msg)1
2.14
\dots 
5 + 4 + 3 + 2 + 1 = 15.
\boxed{\textbf{(D) } 15}.$== Solution 3 ==
We can just count the cases directly since there are so little.
Thus, our answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(D) } 15}.$
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. 
