2019 AIME I Problems/Problem 4

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 $1 + 121 + 310 + 690 = \boxed{122}$ .

2019 AIME I (Problems • Answer Key • Resources)
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2019 AIME I Problems/Problem 4

Contents

Problem 4

Solution 1 (Recursion)

Solution 2 (Casework)

See Also