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

Revision as of 12:47, 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

@@ Line 86: / Line 86: @@
 \end{align*}
 </cmath>
-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>.
+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>.
 ~Tau

2022 AMC 12A (Problems • Answer Key • Resources)
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Preceded by Problem 23	Followed by Problem 25
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