2019 AIME I Problems/Problem 4
The 2019 AIME I takes place on March 13, 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 
be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when is divided by 1000.
Solution
We can perform casework. Call the substitution area the "bench". Case 1: No substitutions. 1.
Case 2: One substitution. Choose one player on the field to sub out, and one player on the bench = 11 * 11 = 121.
Case 3: Two substitutions. Choose one player on the field to sub out, and one player on the bench = 11 * 11 so far. Now choose one player on the field to sub out, and one player on the bench that was not the original player subbed out = * 11 * 10 = 13310 = 310.
Case 4: Three substitutions. Using similar logic as with Case 3 we get (11 * 11) * (11 * 10) * (11 * 9) which ends in 690, so the answer is 1 + 121 + 1000 = 
.
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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