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Difference between revisions of "2019 AIME I Problems/Problem 4"

m (→‎Solution 1 (Recursion))
m (→‎Solution 2 (Casework): Boldfaced cases)
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We can perform casework. Call the substitution area the "bench".
 
We can perform casework. Call the substitution area the "bench".
  
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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{6}</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>. T
  
 
Therefore, the answer is <math>1 + 121 + 310 + 690 = \boxed{122}</math>.
 
Therefore, the answer is <math>1 + 121 + 310 + 690 = \boxed{122}</math>.

Revision as of 11:56, 26 December 2019

Problem 4

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 $1+11^2+11^3\cdot (10+11\cdot9)= 1+11^2+11^3\cdot (1000)$. 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 $11\cdot 11\cdot 11\cdot 10 = 13310\equiv 310 \bmod{6}$.

$\textbf{Case 4}$: 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 $690$. T

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

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

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