Difference between revisions of "2015 AMC 8 Problems/Problem 10"
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How many integers between 1000 and 9999 have four distinct digits? | How many integers between 1000 and 9999 have four distinct digits? | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=9|num-a=11}} | {{AMC8 box|year=2015|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Combinatorics Problems]] | ||
Latest revision as of 19:43, 26 June 2025
Contents
Problem 10
How many integers between 1000 and 9999 have four distinct digits?
Solution 1
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.
If you're too lazy to do 
, we can notice 
and 
The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. 
)
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Nothing
Video Solution
https://youtu.be/Zhsb5lv6jCI?t=272
https://www.youtube.com/watch?v=OESYIYjZFdk ~David
~🐛
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
