Difference between revisions of "2015 AMC 8 Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | + | There are <math>9</math> choices for the first number, since it cannot be <math>0</math>, there are only <math>9</math> choices left for the second number since it must differ from the first, <math>8</math> choices for the third number, since it must differ from the first two, and <math>7</math> choices for the fourth number, since it must differ from all three. This means there are <math>9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}</math> integers between <math>1000</math> and <math>9999</math> with four distinct digits. | |
==Video solution== | ==Video solution== | ||
Revision as of 12:35, 17 December 2022
Contents
Problem
How many integers between 
and 
have four distinct digits?
Solution
There are 
choices for the first number, since it cannot be 
, there are only choices left for the second number since it must differ from the first, 
choices for the third number, since it must differ from the first two, and 
choices for the fourth number, since it must differ from all three. This means there are 
integers between and with four distinct digits.
Video solution
https://youtu.be/Zhsb5lv6jCI?t=272
https://www.youtube.com/watch?v=OESYIYjZFdk
~savannahsolver
See Also
| 2015 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
