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.

1 Problem
2 Solution 1 (Parking Functions)
3 Solution 2 (Casework)
4 Solution 3 (Recursive Equations Approach)
5 Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)
6 Solution 5 (Mod 5 Trick Redeemed)
7 Video Solution by Math-X (Smart and Simple)
8 Video Solution
9 Video Solution
10 Video Solution By OmegaLearn using Complementary Counting
11 See Also

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 equivalent to 1 (mod n). Thus 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

Solution 5 (Mod 5 Trick Redeemed)

Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:

Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by $5$ . Therefore to evaluate the answer modulo $5$ , we only need to look at sets where each element has a multiplicity of at least $2$ (i.e. appears twice or more in the set).

These sets are of the form $\{a,a,b,b,b\}$ and $\{a,a,a,a,a\}$ . The first set can be permuted in $\binom{5}{2,3}=10 \equiv 0\pmod{5}$ , and the second set can be permuted one way, and the only set of the form $\{a,a,a,a,a\}$ is $\{0,0,0,0,0\}$ . Therefore the answer is congruent to $1\pmod{5}$ and you CAN choose $\boxed{\textbf{(E) }1296}$ .

~SpencerD.

Video Solution by Math-X (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=aBXOzKzoAZwkb4l1&t=9314 ~Math-X

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

2022 AMC 10A (Problems • Answer Key • Resources)
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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)

Solution 5 (Mod 5 Trick Redeemed)

Video Solution by Math-X (Smart and Simple)

Video Solution

Video Solution

Video Solution By OmegaLearn using Complementary Counting

See Also