Difference between revisions of "2015 AMC 8 Problems/Problem 10"
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==Video solution== | ==Video solution== | ||
https://youtu.be/Zhsb5lv6jCI?t=272 | https://youtu.be/Zhsb5lv6jCI?t=272 | ||
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| + | https://www.youtube.com/watch?v=OESYIYjZFdk | ||
==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}} | {{MAA Notice}} | ||
Revision as of 18:45, 16 January 2022
Contents
Problem
How many integers between 
and 
have four distinct digits?
Solution
There arechoices for the first number, since it cannot be
, there are only
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.
choices for the first number, since it cannot be 
, there are only 
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 Video solution
https://youtu.be/Zhsb5lv6jCI?t=272
https://www.youtube.com/watch?v=OESYIYjZFdk
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. 
