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 with two <math>1</math>s. We also have to take into account "double <math>1</math>s," namely, <math>112002, 211002, 201102, 200112,</math> and <math>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. | ||
| + | |||
| + | Case 1: The first <math>1</math> is ahead of the first <math>2.</math> | ||
| + | Then the second <math>1</math> has <math>5</math> places. | ||
| + | |||
| + | Case 2: The first <math>1</math> is below the first <math>2.</math> | ||
| + | Then the second <math>1</math> has <math>4</math> places. | ||
| + | |||
| + | <math> \dots </math> | ||
| + | |||
| + | This continues as the answer comes to | ||
| + | |||
| + | <math> 5 + 4 + 3 + 2 + 1 = \frac{5(6)}{2}=3(5)=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}.</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}} | ||
Latest revision as of 20:00, 15 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 with two s. We also have to take into account "double s," namely, 
and 
in our spots.
Thus, our answer is 
Solution 2
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 
Solution 3
We can just count the cases directly since there are so little.
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. 
