2012 AMC 8 Problems/Problem 10: Difference between revisions
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Since the two <math> 2 </math>'s are indistinguishable, there are <math> \frac {3\cdot6}{2} </math> such numbers <math> \implies \boxed{\textbf{(D)}\ 9} </math>. | Since the two <math> 2 </math>'s are indistinguishable, there are <math> \frac {3\cdot6}{2} </math> such numbers <math> \implies \boxed{\textbf{(D)}\ 9} </math>. | ||
==Solution 3(a simpler version of Solution 2)== | |||
==Video Solution== | ==Video Solution== | ||
Revision as of 16:20, 2 January 2023
Problem
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
Solution 1
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of 
, since all of the valid 4-digit number will always be greater than 
. The best way to solve this problem is by using casework.
There can be only two leading digits, namely 
or 
.
When the leading digit is , you can make 
such numbers.
When the leading digit is , you can make 
such numbers.
Summing the amounts of numbers, we find that there are 
such numbers.
Solution 2
Notice that the first digit cannot be 
, as the number is greater than . Therefore, there are three digits that can be in the thousands.
The rest three digits of the number have no restrictions, and therefore there are for each leading digit.
Since the two 's are indistinguishable, there are 
such numbers 
.
Solution 3(a simpler version of Solution 2)
Video Solution
https://youtu.be/OBKIeTgw4Zg ~savannahsolver
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
