Difference between revisions of "2019 AIME I Problems/Problem 4"
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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 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>. | ||
| − | 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 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>. |
~BJHHar | ~BJHHar | ||
Revision as of 20:26, 1 December 2019
Problem 4
A soccer team has 
available players. A fixed set of 
players starts the game, while the other are available as substitutes. During the game, the coach may make as many as 
substitutions, where any one of the 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 
.
Solution 1 (Recursion)
There are 
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 subs is the product of the number of new subs (
) and the players that can be ejected (). The formula for subs is then 
with 
.
Summing from 0 to 3 gives 
. Notice that 
. Then, rearrange it into 
. When taking modulo 1000, the last term goes away. What is left is 
.
~BJHHar
Solution 2 (Casework)
We can perform casework. Call the substitution area the "bench".
Case 1: No substitutions. There is 
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 
.
Case 3: Two substitutions. Choose one player on the field to sub out, and one player on the bench. Once again, this is 
. 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 
.
Case 4: Three substitutions. Using similar logic as Case 3, we get 
. The resulting number ends in 
. T
Therefore, the answer is 
.
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. 
