Difference between revisions of "2022 AMC 10A Problems/Problem 24"

(Video Solution By OmegaLearn using Complementary Counting)
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<math>\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296</math>
 
<math>\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296</math>
  
== Video Solution By OmegaLearn using Complementary Counting ==  
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== Solution 1 By OmegaLearn using Complementary Counting ==  
  
 
https://youtu.be/jWoxFT8hRn8
 
https://youtu.be/jWoxFT8hRn8
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
== See Also ==
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{{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}
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{{MAA Notice}}
  
 
== See Also ==
 
== See Also ==

Revision as of 03:12, 12 November 2022

Problem

How many strings of length $5$ formed from the digits $0$,$1$,$2$,$3$,$4$ are there such that for each $j\in\{1,2,3,4\}$, at least $j$ of the digits are less than $j$? (For example, $02214$ satisfies the 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 $2$.)

$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$

Solution 1 By OmegaLearn using Complementary Counting

https://youtu.be/jWoxFT8hRn8

~ pi_is_3.14

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png