2022 AMC 10A Problem 24

Attempt Problem 24 of the 2022 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

How many strings of length 55 formed from the digits 0,0, 1,1, 2,2, 3,3, 4,4, are there such that for each j∈{1,2,3,4},j \in \{1,2,3,4\}, at least jj of the digits are less than j?j?
(For example, 0221402214 satisfies this condition because it contains at least 11 digit less than 1,1, at least 22 digits less than 2,2, at least 33 digits less than 3,3, and at least 44 digits less than 4.4. The string 2340423404 does not satisfy the condition because it does not contain at least 22 digits less than 2.2.)
500500
625625
10891089
11991199
12961296
Answer: E
Concepts:parking functionscircular countingarrangements with restrictions
Small hint:
Interpret each digit as a car’s preferred parking space
Big hint:
Add a sixth parking space and arrange the spaces in a circle
Solution:
Regard the five digits, in order, as the preferred parking spaces of five cars. Spaces are numbered 0,1,2,3,4,0,1,2,3,4, and each car takes its preferred space if possible, or else the first empty space to its right. If the preferences sorted into nondecreasing order are b1≤b2≤⋯≤b5,b_1\le b_2\le\cdots\le b_5, all cars park exactly when bi≤i−1(1≤i≤5).b_i\le i-1\qquad(1\le i\le5). These inequalities are precisely the conditions in the problem.
To count such preference strings, add a sixth space and arrange spaces 0,1,…,50,1,\ldots,5 in a circle. For any of the 656^5 preference strings, all five cars park and exactly one space remains empty. Rotating every preference by one position rotates the empty space as well. Thus each orbit of six preference strings has each possible empty space exactly once.
Therefore exactly 656=64=1296\frac{6^5}{6}=6^4=1296 circular preference strings leave space 55 empty. No car in such a string prefers space 5,5, and cutting the circle immediately after that empty space gives exactly a successful parking sequence on spaces 00 through 4.4. Hence the desired number of strings is 1296.1296.
Thus, E is the correct answer.
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