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

(Video Solution By OmegaLearn using Complementary Counting)
Line 4: Line 4:
 
<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 ==  
+
== 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 ==
 +
 +
{{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}
 +
{{MAA Notice}}
  
 
== See Also ==
 
== See Also ==

Revision as of 04:12, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png