2022 AMC 10A Problems/Problem 24

The following problem is from both the 2022 AMC 10A #24 and 2022 AMC 12A #24, so both problems redirect to this page.

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 1 (Parking Functions)

For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.

Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$ , we can add $1$ to each $c_i$ (mod $n+1$ ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.

Plugging in $n = 5$ gives $\boxed{\textbf{(E) }1296}$ such strings.

~oh54321

Solution 2 (Casework)

Note that a valid string must have at least one $0.$

We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:

The string has $1$ different digit.

The only possibility is $00000.$

There is $\boldsymbol{1}$ string in this case.

The string has $2$ different digits.

We have the following table: $\begin{array}{c||c|c|c|c||c} & & & & & \\ [-2.5ex] \textbf{Digits} & \boldsymbol{01} & \boldsymbol{02} & \boldsymbol{03} & \boldsymbol{04} & \textbf{Row's Count} \\ [0.5ex] \hline & & & & & \\ [-1.5ex] & 00001 & 00002 & 00003 & 00004 & \hspace{2mm}4\cdot\frac{5!}{4!1!}=20 \\ [2ex] & 00011 & 00022 & 00033 & & \hspace{2mm}3\cdot\frac{5!}{3!2!}=30 \\ [2ex] & 00111 & 00222 & & & \hspace{2mm}2\cdot\frac{5!}{2!3!}=20 \\ [2ex] & 01111 & & & & 1\cdot\frac{5!}{1!4!}=5 \\ [0.75ex] \end{array}$ There are $\boldsymbol{20+30+20+5=75}$ strings in this case.

The string has $3$ different digits.

We have the following table: $\begin{array}{c||c|c|c|c|c|c||c} & & & & & & & \\ [-2.5ex] \textbf{Digits} & \boldsymbol{012} & \boldsymbol{013} & \boldsymbol{014} & \boldsymbol{023} & \boldsymbol{024} & \boldsymbol{034} & \textbf{Row's Count} \\ [0.5ex] \hline & & & & & & & \\ [-1.5ex] & 00012 & 00013 & 00014 & 00023 & 00024 & 00034 & \hspace{2mm}6\cdot\frac{5!}{3!1!1!}=120 \\ [2ex] & 00112 & 00113 & 00114 & 00223 & 00224 & & \hspace{2mm}5\cdot\frac{5!}{2!2!1!}=150 \\ [2ex] & 00122 & 00133 & & 00233 & & & 3\cdot\frac{5!}{2!1!2!}=90 \\ [2ex] & 01112 & 01113 & 01114 & & & & 3\cdot\frac{5!}{1!3!1!}=60 \\ [2ex] & 01122 & 01133 & & & & & 2\cdot\frac{5!}{1!2!2!}=60 \\ [2ex] & 01222 & & & & & & 1\cdot\frac{5!}{1!1!3!}=20 \\ [0.75ex] \end{array}$ There are $\boldsymbol{120+150+90+60+60+20=500}$ strings in this case.

The string has $4$ different digits.

We have the following table: $\begin{array}{c||c|c|c|c} & & & & \\ [-2.5ex] \textbf{Digits} & \boldsymbol{0123} & \boldsymbol{0124} & \boldsymbol{0134} & \boldsymbol{0234} \\ [0.5ex] \hline & & & & \\ [-1.5ex] & 00123 & 00124 & 00134 & 00234 \\ [2ex] & 01123 & 01124 & 01134 & \\ [2ex] & 01223 & 01224 & & \\ [2ex] & 01233 & & & \\ [0.75ex] \end{array}$ There are $\boldsymbol{10\cdot\frac{5!}{2!1!1!1!}=600}$ strings in this case.

The string has $5$ different digits.

There are $\boldsymbol{5!=120}$ strings in this case.

Together, the answer is $1+75+500+600+120=\boxed{\textbf{(E) }1296}.$

~MRENTHUSIASM

Solution 3 (Recursive Equations Approach)

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 4 (Fake solve, incorrect logic, correct answer by coincidence)

The number of strings is $(n+1)^(n-1)$ as shown by Solution 1 (Parking Function), which is always equal to. L1 (mod n). This you can choose $\boxed{\textbf{(E) }1296}$ .

However, you **CANNOT** prove that fact from this *incorrect* argument:

Let the set of all valid sequences be $S$ . Notice that for any sequence $\{a_1,a_2,a_3,a_4,a_5\}$ in $S$ , the 5 cyclic permutations $\begin{align*} \{a_1,a_2,a_3,a_4,a_5\}\\ \{a_2,a_3,a_4,a_5,a_1\}\\ \{a_3,a_4,a_5,a_1,a_2\}\\ \{a_4,a_5,a_1,a_2,a_3\}\\ \{a_5,a_1,a_2,a_3,a_4\} \end{align*}$ must also belong in $S$ . However, one must consider the edge case all 5 elements are the same (only $\{0,0,0,0,0\}$ ), in which case all sequences listed are equivalent.

Alas, one must also consider cases like 10101 that only have 3 distinct cyclic permutations, so we cannot get a useful constraint from this view, unless *all* the cases are categorized and counted.

Thus it is NOT implied that $\lvert S \rvert \equiv 1 \pmod 5$ , so it is not justified to choose $\boxed{\textbf{(E) }1296}$ by inspection of answer options.

If you solve $S(n)$ for smaller $n$ by bashing, you can find $|S(1)| =1$ , $|S(2)| = 3$ , $|S(3)| =16$ , $|S(4)| =125$ , which leads to an Engineer's Induction guess that $|S(n)| \equiv 1 pmod n$ , or the exact formula $(n+1)^(n-1)$

~ idea by Tau, logic corrected by oinava

Video Solution

https://youtu.be/130OKAfG_-o

~MathProblemSolvingSkills.com

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

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2022 AMC 10A Problems/Problem 24

Contents

Problem

Solution 1 (Parking Functions)

Solution 2 (Casework)

Solution 3 (Recursive Equations Approach)

Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)

Video Solution

Video Solution

Video Solution By OmegaLearn using Complementary Counting

See Also