Difference between revisions of "2022 AMC 12A Problems/Problem 24"
Pi is 3.14 (talk | contribs) (→See Also) |
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | == Video Solution By ThePuzzlr == | ||
+ | https://youtu.be/qWIYzgqgCNg | ||
+ | |||
+ | ~ MathIsChess | ||
==Video Solution== | ==Video Solution== |
Revision as of 09:46, 12 November 2022
Contents
[hide]Problem
How many strings of length 5 formed from the digits 0, 1, 2, 3, 4 are there such that for each , at least of the digits are less than ? (For example, 02214 satisfies this condition because it contains at least 1 digit less than 1, at least 2 digits less than 2, at least 3 digits less than 3, and at least 4 digits less than 4. The string 23404 does not satisfy the condition because it does not contain at least 2 digits less than 2.)
Solution
Denote by the number of -digit strings formed by using numbers , where for each , at least of the digits are less than .
We have the following recursive equation: and the boundary condition for any .
By solving this recursive equation, for and , we get
For and , we get
For and , we get
For and , we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By ThePuzzlr
https://youtu.be/qWIYzgqgCNg
~ MathIsChess
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By OmegaLearn using Complementary Counting
~ pi_is_3.14
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 AMC 12 Problems and Solutions |
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.