Difference between revisions of "2022 AMC 12A Problems/Problem 24"

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution 2 (Cheese)==
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Let the set of all valid sequences be <math>S</math>.
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Notice that for any sequence <math>\{a,b,c,d,e\}</math> in <math>S</math>, the sequences
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<cmath>
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\begin{align*}
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\{b,c,d,e,a\}\
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\{c,d,e,a,b\}\
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\{d,e,a,b,c\}\
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\{e,a,b,c,d\}
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\end{align*}
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</cmath>
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must also belong in <math>S</math>. The only exception is when all 5 elements are the same, i.e. <math>\{0,0,0,0,0\}</math>. Then <math>\lvert S \rvert \equiv 1 ( \pmod 5)</math>, which by inspection of the answer choices yields <math>\boxed{(E) 1296}</math>.
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~Tau
  
 
== Video Solution By ThePuzzlr ==  
 
== Video Solution By ThePuzzlr ==  

Revision as of 12:46, 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 this 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

Denote by $N \left( p, q \right)$ the number of $p$-digit strings formed by using numbers $0, 1, \cdots, q$, where for each $j \in \{1,2, \cdots , q\}$, at least $j$ of the digits are less than $j$.

We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1\] and the boundary condition $N \left( p, 0 \right) = 1$ for any $p \geq 0$.

By solving this recursive equation, for $q = 1$ and $p \geq q$, we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}

For $q = 2$ and $p \geq q$, we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}

For $q = 3$ and $p \geq q$, we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}

For $q = 4$ and $p = 5$, we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{\textbf{(E) 1296}}  . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2 (Cheese)

Let the set of all valid sequences be $S$. Notice that for any sequence $\{a,b,c,d,e\}$ in $S$, the sequences \begin{align*} \{b,c,d,e,a\}\\ \{c,d,e,a,b\}\\ \{d,e,a,b,c\}\\ \{e,a,b,c,d\} \end{align*} must also belong in $S$. The only exception is when all 5 elements are the same, i.e. $\{0,0,0,0,0\}$. Then $\lvert S \rvert \equiv 1 ( \pmod 5)$, which by inspection of the answer choices yields $\boxed{(E) 1296}$.

~Tau

Video Solution By ThePuzzlr

https://youtu.be/qWIYzgqgCNg

~ MathIsChess

Video Solution

https://youtu.be/mj78e_LnkX0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution By OmegaLearn using Complementary Counting

https://youtu.be/jWoxFT8hRn8

~ pi_is_3.14

See Also

2022 AMC 12A (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 12 Problems and Solutions
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

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