2015 AMC 8 Problems/Problem 10: Difference between revisions
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==Video solution== | ==Video solution== | ||
https://youtu.be/Zhsb5lv6jCI?t=272 | https://youtu.be/Zhsb5lv6jCI?t=272 | ||
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
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 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
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.