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

Latest revision as of 16:08, 31 October 2024

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 Solution 6 (Casework Bash)
8 Solution 7 (Complementary Counting)
9 Video Solution by Math-X (Smart and Simple)
10 Video Solution
11 Video Solution
12 Video Solution By OmegaLearn using Complementary Counting
13 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

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:

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 $5$ (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. 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

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

Solution 7 (Complementary Counting)

Let us call the condition where you require at least $j$ digits less than $j$ Rule $j$ . For a sequence to break Rule $1$ , it must contain no $0$ s. There are $4^5=1024$ ways to do this. For a sequence to break Rule $2$ but not Rule $1$ , it must contain exactly 1 $0$ . However, this one $0$ contributes toward completing the rule, meaning all other digits must be greater than $2$ . After placing the one $0$ in any of $5$ slots, there are $4$ slots left with $3$ options each, giving a total of $5 \cdot 3^4 = 405$ ways. For a sequence to break Rule $3$ but neither Rule $2$ or $1$ , it must have two digits less than $2$ (either $(0, 0)$ or $(0, 1)$ ) and all other digits must be $3$ or $4$ . There are a total of $30$ ways to arrange the first two digits and $2^3$ ways to arrange the latter three, leaving $240$ ways to do this. Finally, for a sequence to only break rule $4$ , it must have three digits less than $3$ . They could be $(0, 1, 2)$ (60 arr.), $(0, 1, 1); (0, 0, 1); (0, 0, 2)$ ( $30$ arr. each), or $(0, 0, 0)$ ( $10$ arr.) This gives $160$ possible arrangements overall.

There are in total $5^5=3125$ possible arrangements, so by adding all of the invalid states together and subtracting from $3125$ , we get our answer, which is $\boxed{\textbf{(E) }1296}$

~ Shadowleafy

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

@@ Line 20: / Line 20: @@
 Note: If you are confused, the problem is equivalent to “How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math> are there such that for each <math>j \in \{1,2,3,4,5\}</math>, at least <math>j</math> of the digits are less than <math>j</math>?”
 ~inaccessibles
@@ Line 185: / Line 186: @@
 ==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.
 - 1
@@ Line 269: / Line 269: @@
 - 120
 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)
@@ Line 275: / Line 276: @@
 ~ MathHayden
+==Solution 7 (Complementary Counting)==
+Let us call the condition where you require at least <math>j</math> digits less than <math>j</math> Rule <math>j</math>.
+For a sequence to break Rule <math>1</math>, it must contain no <math>0</math>s. There are <math>4^5=1024</math> ways to do this.
+For a sequence to break Rule <math>2</math> but not Rule <math>1</math>, it must contain exactly 1 <math>0</math>. However, this one <math>0</math> contributes toward completing the rule, meaning all other digits must be greater than <math>2</math>. After placing the one <math>0</math> in any of <math>5</math> slots, there are <math>4</math> slots left with <math>3</math> options each, giving a total of <math>5 \cdot 3^4 = 405</math> ways.
+For a sequence to break Rule <math>3</math> but neither Rule <math>2</math> or <math>1</math>, it must have two digits less than <math>2</math> (either <math>(0, 0)</math> or <math>(0, 1)</math>) and all other digits must be <math>3</math> or <math>4</math>. There are a total of <math>30</math> ways to arrange the first two digits and <math>2^3</math> ways to arrange the latter three, leaving <math>240</math> ways to do this.
+Finally, for a sequence to only break rule <math>4</math>, it must have three digits less than <math>3</math>. They could be <math>(0, 1, 2)</math> (60 arr.), <math>(0, 1, 1); (0, 0, 1); (0, 0, 2)</math> (<math>30</math> arr. each), or <math>(0, 0, 0)</math> (<math>10</math> arr.) This gives <math>160</math> possible arrangements overall.
+There are in total <math>5^5=3125</math> possible arrangements, so by adding all of the invalid states together and subtracting from <math>3125</math>, we get our answer, which is <math>\boxed{\textbf{(E) }1296}</math>
+~ Shadowleafy
 ==Video Solution by Math-X (Smart and Simple)==

2022 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AMC 10A Problems/Problem 24"

Latest revision as of 16:08, 31 October 2024

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)

Solution 6 (Casework Bash)

Solution 7 (Complementary Counting)

Video Solution by Math-X (Smart and Simple)

Video Solution

Video Solution

Video Solution By OmegaLearn using Complementary Counting

See Also