Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1 (Parking Functions))
(→‎Solution 6 (Casework Bash))
Line 186: Line 186:
 
==Solution 6 (Casework Bash)==
 
==Solution 6 (Casework Bash)==
 
We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string.
 
We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string.
−
 
+
[hide]00000 - 1
−
00000 - 1
 
  
 
00001 - 5
 
00001 - 5
Line 269: Line 268:
 
01233 - 60
 
01233 - 60
  
−
01234 - 120
+
01234 - 120[/hide]
 +
 
  
 
Finally, adding all of these up, we obtain <math>1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{\textbf{(E) }1296}</math>. (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)
 
Finally, adding all of these up, we obtain <math>1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{\textbf{(E) }1296}</math>. (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)

Revision as of 18:58, 17 August 2024

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

Note: If you are confused, the problem is equivalent to “How many strings of length $5$ formed from the digits $0$, $1$, $2$, $3$, $4$, $5$ are there such that for each $j \in \{1,2,3,4,5\}$, at least $j$ of the digits are less than $j$?”

~inaccessibles

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:

  1. The string has $1$ different digit.

    2. The only possibility is $00000.$

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

  2. The string has $2$ different digits.

    3. 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.

  3. The string has $3$ different digits.

    4. 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.

  4. The string has $4$ different digits.

    5. 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.

  5. The string has $5$ different digits.

    6. 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 $5$ (alternatively there is no factor of 5 in the denominator of the multinomial coefficient unless an element is repeated 5 times). 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.

Solution 6 (Casework Bash)

We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string. [hide]00000 - 1

00001 - 5

00002 - 5

00003 - 5

00004 - 5

00011 - 10

00012 - 20

00013 - 20

00014 - 20

00022 - 10

00023 - 20

00024 - 20

00033 - 10

00034 - 20

00111 - 10

00112 - 30

00113 - 30

00114 - 30

00122 - 30

00123 - 60

00124 - 60

00133 - 30

00134 - 60

00222 - 10

00223 - 30

00224 - 30

00233 - 30

00234 - 60

01111 - 5

01112 - 20

01113 - 20

01114 - 20

01122 - 30

01123 - 60

01124 - 60

01133 - 30

01134 - 60

01222 - 20

01223 - 60

01224 - 60

01233 - 60

01234 - 120[/hide]


Finally, adding all of these up, we obtain $1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{\textbf{(E) }1296}$. (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)

The key to organizing all of these cases is to increase the digits one by one, starting from the right and iterating through all the possible cases.

~ MathHayden

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

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
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
2022 AMC 12A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_24&oldid=225218"