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 12A Problems/Problem 19"

(→‎Solution)
(→‎Solution 2)
Line 74: Line 74:
  
 
==Solution 2==
 
==Solution 2==
−
To solve this problem, we can use recursion on <math>n</math>. Let <math>A_n</math> be the number of arrangement for <math>n</math> numbers. Now, let's look at how these arrangements are formed by case work on the first number <math>a_1</math>.
+
To solve this problem, we can use recursion on <math>n</math>. Let <math>A_n</math> be the number of arrangements for <math>n</math> numbers. Now, let's look at how these arrangements are formed by case work on the first number <math>a_1</math>.
  
−
If <math>a_1 = 1</math>, the remaining <math>n-1</math> numbers from <math>2</math> to <math>n</math> are arranged in the same way just like number 1 to <math>n-1</math>. So there are <math>A_{n-1}</math> arrangements.
+
If <math>a_1 = 1</math>, the remaining <math>n-1</math> numbers from <math>2</math> to <math>n</math> are arranged in the same way just like number 1 to <math>n-1</math> in the case of <math>n-1</math> numbers. So there are <math>A_{n-1}</math> arrangements.
  
 
If <math>a_1 = 2</math>, then we need to choose 1 position from position 2 to <math>n-1</math> to put 1, and all remaining numbers must be arranged in increasing order, so there are <math>\binom{n-1}{1}</math> such arrangements.
 
If <math>a_1 = 2</math>, then we need to choose 1 position from position 2 to <math>n-1</math> to put 1, and all remaining numbers must be arranged in increasing order, so there are <math>\binom{n-1}{1}</math> such arrangements.

Revision as of 10:27, 13 November 2022

Problem

Suppose that 13 cards numbered $1, 2, 3, \cdots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards 1, 2, 3 are picked up on the first pass, 4 and 5 on the second pass, 6 on the third pass, 7, 8, 9, 10 on the fourth pass, and 11, 12, 13 on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?

[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$

Solution

Solution 1

Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$. Also note that if $m$, which is the card that is numbered one more than $n$, is placed before $n$, then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.

To construct this, $n$ cannot go in the $n$th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$. Therefore $n$ must be in slots $n+1$ to $13$. Let's do casework on which slot $n$ goes into to get a general idea for how the problem works.


$\textbf{Case 1:}$ With $n$ in spot $n+1$, there are $n$ available slots before $n$, and there are $n-1$ cards preceding $n$. Therefore the number of ways to reserve these slots for the $n-1$ cards is $\binom{n}{n-1}$. Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$, giving only $1$ way. Therefore in this case there are $\binom{n}{n-1}$ possibilities.


$\textbf{Case 2:}$ With $n$ in spot $n+2$, there are $n+1$ available slots before $n$, and there are $n-1$ cards preceding $n$. Therefore the number of ways to reserve slots for these cards are $\binom{n+1}{n-1}$. Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$, which yields $1$ way. Therefore, this case has $\binom{n+1}{n-1}$ possibilities.



I think we can see a general pattern now. With $n$ in slot $x$, there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$, this sum amounts to: \[\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\] for any $n$.

Hmmm... where have we seen this before?

We use wishful thinking to add a term of $\binom{n-1}{n-1}$: \[\binom{n-1}{n-1}+\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\]

This is just the hockey stick identity! Applying it, this expression is equal to $\binom{13}{n}$. However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is \[\binom{13}{n}-1\]

Finally, to calculate the total for all $n$, we sum from $n=0$ to $13$. This yields us:

\[\sum_{n=0}^{13} \binom{13}{n}-1 \implies \sum_{n=0}^{13} \binom{13}{n} - \sum_{n=0}^{13} 1\] \[\implies 2^{13} - 14 = 8192 - 14 = 8178 = \boxed{D}\]


~KingRavi


Solution 2

To solve this problem, we can use recursion on $n$. Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$.

If $a_1 = 1$, the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.

If $a_1 = 2$, then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{1}$ such arrangements.

If $a_1 = k$, then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\cdots k-1$, and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{k-1}$ such arrangements.

So we can write \[A_n = A_{n-1} + \binom{n-1}{1} + \binom{n-1}{2} + \cdots + \binom{n-1}{n-1}\] which can be simplified to \[A_n = A_{n-1} + 2^{n-1} - 1\] We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula \[A_n - A_{n-1} = 2^{n-1} - 1\] \[A_{n-1} - A_{n-2} = 2^{n-2} - 1\] \[\cdots\] \[A_2 - A_{1} = 2^{1} - 1\] to get \[A_n - A_1 = \sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1\] since $A_1=0$, we have \[A_n = 2^n - n - 1\] and $A_{13} = 2^{13} - 14 = \boxed{8178}$.

So the answer is $\boxed{D}$

-- Dan Li

Video Solution By ThePuzzlr

https://youtu.be/p9xNduqTKLM

~ MathIsChess

Solution by OmegaLearn Using Combinatorial Identities and Overcounting

https://youtu.be/gW8gPEEHSfU

~ pi_is_3.14

Solution

https://youtu.be/ZGqrs5eg6-s

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

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_12A_Problems/Problem_19&oldid=181280"