2022 AMC 10A Problems/Problem 24
- The following problem is from both the 2022 AMC 10A #24 and 2022 AMC 12A #24, so both problems redirect to this page.
Contents
Problem
How many strings of length formed from the digits , , , , are there such that for each , at least of the digits are less than ? (For example, satisfies this condition because it contains at least digit less than , at least digits less than , at least digits less than , and at least digits less than . The string does not satisfy the condition because it does not contain at least digits less than .)
Solution 1 (Parking Functions)
For some , let there be parking spaces counterclockwise in a circle. Consider a string of integers each between and , and let cars come into this circle so that the th car tries to park at spot , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of numbers between and that contain at least integers for are exactly the set of strings that leave spot empty. Also note for any string , we can add to each (mod ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one with so that leaves spot empty. This gives there are such strings.
Plugging in gives such strings.
~oh54321
Solution 2 (Casework)
Note that a valid string must have at least one
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
- The string has different digit.
- The string has different digits.
- The string has different digits.
- The string has different digits.
- The string has different digits.
The only possibility is
There is string in this case.
We have the following table: There are strings in this case.
We have the following table: There are strings in this case.
We have the following table: There are strings in this case.
There are strings in this case.
Together, the answer is
~MRENTHUSIASM
Solution 3 (Recursive Equations Approach)
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)
Solution 4 (Answer Choices)
Let the set of all valid sequences be . Notice that for any sequence in , the sequences must also belong in . However, one must consider the edge case all 5 elements are the same (only ), in which case all sequences listed are equivalent. Then , which yields by inspection.
~Tau
Video Solution
~MathProblemSolvingSkills.com
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 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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.