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 "2019 AIME I Problems/Problem 4"

m (→‎Solution 1 (Recursion))
(→‎Solution 2 (Casework))
 
(13 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
==Problem 4==
+
==Problem==
 
A soccer team has <math>22</math> available players. A fixed set of <math>11</math> players starts the game, while the other <math>11</math> are available as substitutes. During the game, the coach may make as many as <math>3</math> substitutions, where any one of the <math>11</math> players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let <math>n</math> be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when <math>n</math> is divided by <math>1000</math>.
 
A soccer team has <math>22</math> available players. A fixed set of <math>11</math> players starts the game, while the other <math>11</math> are available as substitutes. During the game, the coach may make as many as <math>3</math> substitutions, where any one of the <math>11</math> players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let <math>n</math> be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when <math>n</math> is divided by <math>1000</math>.
  
 
==Solution 1 (Recursion)==
 
==Solution 1 (Recursion)==
−
There are <math>0-3</math> substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for 0 subs is 1, and the ways to reorganize after <math>n</math> subs is the product of the number of new subs (<math>12-n</math>) and the players that can be ejected (<math>11</math>). The formula for <math>n</math> subs is then <math>a_n=11(12-n)a_{n-1}</math> with <math>a_0=1</math>.  
+
There are <math>0-3</math> substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for <math>0</math> subs is <math>1</math>, and the ways to reorganize after <math>n</math> subs is the product of the number of new subs (<math>12-n</math>) and the players that can be ejected (<math>11</math>). The formula for <math>n</math> subs is then <math>a_n=11(12-n)a_{n-1}</math> with <math>a_0=1</math>.  
  
−
Summing from 0 to 3 gives <math>1+11^2+11^{3}10+11^{4}10\cdot9</math>. Notice that <math>10+9\cdot11\cdot10=10+990=1000</math>. Then, rearrange it into <math>1+11^2+11^3(10+11\cdot9)=1+11^2+11^3(1000)</math>. When taking modulo 1000, the last term goes away. What is left is <math>1+11^2=\boxed{122}</math>.
+
Summing from <math>0</math> to <math>3</math> gives <math>1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9</math>. Notice that <math>10+9\cdot11\cdot10=10+990=1000</math>. Then, rearrange it into <math>1+11^2+11^3\cdot (10+11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)</math>. When taking modulo <math>1000</math>, the last term goes away. What is left is <math>1+11^2=\boxed{122}</math>.
  
 
~BJHHar
 
~BJHHar
Line 12: Line 12:
 
We can perform casework. Call the substitution area the "bench".
 
We can perform casework. Call the substitution area the "bench".
  
−
Case 1: No substitutions. There is <math>1</math> way of doing this: leaving everybody on the field.
+
<math>\textbf{Case 1}</math>: No substitutions. There is <math>1</math> way of doing this: leaving everybody on the field.
  
−
Case 2: One substitution. Choose one player on the field to sub out, and one player on the bench. This corresponds to <math>11\cdot 11 = 121</math>.
+
<math>\textbf{Case 2}</math>: One substitution. Choose one player on the field to sub out, and one player on the bench. This corresponds to <math>11\cdot 11 = 121</math>.
  
−
Case 3: Two substitutions. Choose one player on the field to sub out, and one player on the bench. Once again, this is <math>11\cdot 11</math>. Now choose one more player on the field to sub out and one player on the bench that was not the original player subbed out. This gives us a total of <math>11\cdot 11\cdot 11\cdot 10 = 13310\equiv 310 \bmod{6}</math>.
+
<math>\textbf{Case 3}</math>: Two substitutions. Choose one player on the field to sub out, and one player on the bench. Once again, this is <math>11\cdot 11</math>. Now choose one more player on the field to sub out and one player on the bench that was not the original player subbed out. This gives us a total of <math>11\cdot 11\cdot 11\cdot 10 = 13310\equiv 310 \bmod{1000}</math>.
  
−
Case 4: Three substitutions. Using similar logic as Case 3, we get <math>(11\cdot 11)\cdot (11\cdot 10)\cdot (11\cdot 9)</math>. The resulting number ends in <math>690</math>. T
+
<math>\textbf{Case 4}</math>: Three substitutions. Using similar logic as <math>\textbf{Case 3}</math>, we get <math>(11\cdot 11)\cdot (11\cdot 10)\cdot (11\cdot 9)</math>. The resulting number ends in <math>690</math>.
  
−
Therefore, the answer is <math>1 + 121 + 310 + 690 = \boxed{122}</math>.
+
Therefore, the answer is <math>1 + 121 + 310 + 690 \equiv \boxed{122} \pmod{1000}</math>.
 +
 
 +
~minor mistake fixed by [[User:Prism_melody|Prism Melody]]
 +
 
 +
==Solution 3 (Official MAA)==
 +
There is <math>1</math> way of making no substitutions to the starting lineup. If the coach makes exactly one substitution, this can be done in <math>11^2</math> ways. Two substitutions can happen in <math>11^2\cdot11\cdot 10</math> ways. Similarly, three substitutions can happen <math>11^2\cdot11\cdot10\cdot11\cdot9</math> ways. The total number of possibilities is <math>1+11^2+11^2\cdot11\cdot10+11^2\cdot11\cdot10\cdot11\cdot9=122+11^3(10+990)\equiv 122\pmod{1000}.</math>
 +
 
 +
==Video Solution #1(A bit of casework, a bit of counting)==
 +
https://youtu.be/JQdad7APQG8?t=981
 +
 
 +
==Video Solution==
 +
https://youtu.be/I-8xZGhoDUY
 +
 
 +
~Shreyas S
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2019|n=I|num-b=3|num-a=5}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:32, 5 February 2022

Problem

A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let $n$ be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when $n$ is divided by $1000$.

Solution 1 (Recursion)

There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$, and the ways to reorganize after $n$ subs is the product of the number of new subs ($12-n$) and the players that can be ejected ($11$). The formula for $n$ subs is then $a_n=11(12-n)a_{n-1}$ with $a_0=1$.

Summing from $0$ to $3$ gives $1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9$. Notice that $10+9\cdot11\cdot10=10+990=1000$. Then, rearrange it into $1+11^2+11^3\cdot (10+11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)$. When taking modulo $1000$, the last term goes away. What is left is $1+11^2=\boxed{122}$.

~BJHHar

Solution 2 (Casework)

We can perform casework. Call the substitution area the "bench".

$\textbf{Case 1}$: No substitutions. There is $1$ way of doing this: leaving everybody on the field.

$\textbf{Case 2}$: One substitution. Choose one player on the field to sub out, and one player on the bench. This corresponds to $11\cdot 11 = 121$.

$\textbf{Case 3}$: Two substitutions. Choose one player on the field to sub out, and one player on the bench. Once again, this is $11\cdot 11$. Now choose one more player on the field to sub out and one player on the bench that was not the original player subbed out. This gives us a total of $11\cdot 11\cdot 11\cdot 10 = 13310\equiv 310 \bmod{1000}$.

$\textbf{Case 4}$: Three substitutions. Using similar logic as $\textbf{Case 3}$, we get $(11\cdot 11)\cdot (11\cdot 10)\cdot (11\cdot 9)$. The resulting number ends in $690$.

Therefore, the answer is $1 + 121 + 310 + 690 \equiv \boxed{122} \pmod{1000}$.

~minor mistake fixed by Prism Melody

Solution 3 (Official MAA)

There is $1$ way of making no substitutions to the starting lineup. If the coach makes exactly one substitution, this can be done in $11^2$ ways. Two substitutions can happen in $11^2\cdot11\cdot 10$ ways. Similarly, three substitutions can happen $11^2\cdot11\cdot10\cdot11\cdot9$ ways. The total number of possibilities is $1+11^2+11^2\cdot11\cdot10+11^2\cdot11\cdot10\cdot11\cdot9=122+11^3(10+990)\equiv 122\pmod{1000}.$

Video Solution #1(A bit of casework, a bit of counting)

https://youtu.be/JQdad7APQG8?t=981

Video Solution

https://youtu.be/I-8xZGhoDUY

~Shreyas S

See Also

2019 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2019_AIME_I_Problems/Problem_4&oldid=170969"