2022 AMC 10A Problem 24

Below is the professionally curated solution for Problem 24 of the 2022 AMC 10A, 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.

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Concepts:arrangements with restrictionsmultiset permutationscasework

Difficulty rating: 2390

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

Solution:

Note that there must be at least one 00 to satisfy the condition. We can proceed by casework on the number of distinct digits in the string.

11 digit

The only possible digit is just 0.0. This can be done in 11 way.

22 digits

This gives us a total of 20+30+20+5=75 20 + 30 + 20 + 5 = 75 strings.

33 digits

This gives us a total of 120+150+90+60+60 120 + 150 + 90 + 60 + 60 +20=500+ 20 = 500 strings.

44 digits

This gives us a total of 105!2!1!1!1!=600 10 \cdot \dfrac{5!}{2! \cdot 1! \cdot 1! \cdot 1!} = 600 strings.

55 digits

This gives us 5!=1205! = 120 strings.

All together, we have 1+75+500+600 1 + 75 + 500 + 600 +120=1296+ 120 = 1296 strings.

Thus, E is the correct answer.

Problem 24 in Other Years