Difference between revisions of "2012 AIME I Problems/Problem 3"
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We only need to figure out the number of ways to order the string <math>BBBCCCFFF</math>, where exactly one <math>B</math> is in the first three positions, one <math>C</math> is in the <math>4^{th}</math> to <math>6^{th}</math> positions, and one <math>F</math> is in the last three positions. There are <math>3^3=27</math> ways to place the first <math>3</math> meals. Then for the other two people, there are <math>2</math> ways to serve their meals. Thus, there are <math>(3\cdot2)^3=\boxed{216}</math> ways to serve their meals. | We only need to figure out the number of ways to order the string <math>BBBCCCFFF</math>, where exactly one <math>B</math> is in the first three positions, one <math>C</math> is in the <math>4^{th}</math> to <math>6^{th}</math> positions, and one <math>F</math> is in the last three positions. There are <math>3^3=27</math> ways to place the first <math>3</math> meals. Then for the other two people, there are <math>2</math> ways to serve their meals. Thus, there are <math>(3\cdot2)^3=\boxed{216}</math> ways to serve their meals. | ||
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+ | Note: This solution is wholly wrong, whoever wrote this solution doesn't even understand the problem. | ||
== Video Solution by Richard Rusczyk === | == Video Solution by Richard Rusczyk === |
Revision as of 18:49, 6 March 2021
Contents
Problem
Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.
Solution 1
Call a beef meal a chicken meal and a fish meal Now say the nine people order meals respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by to account for the different ways in which the person to receive the correct meal could be picked. Note, this implies that the dishes are indistinguishable, though the people aren't. For example, two people who order chicken are separate, though if they receive fish, there is only 1 way to order them.
The problem we must solve is to distribute meals to orders with no matches. The two people who ordered 's can either both get 's, both get 's, or get one and one We proceed with casework.
- If the two people both get 's, then the three meals left to distribute must all go to the people. The people then get in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the meals to the people, and there is only 1 way to order this, as all three meals are the same.
- If the two people both get 's, the situation is identical to the above and three possibilities arise.
- If the two people get in some order, then the people must get and the people must get This gives possibilities.
Summing across the cases we see there are possibilities, so the answer is
Solution 2
We only need to figure out the number of ways to order the string , where exactly one is in the first three positions, one is in the to positions, and one is in the last three positions. There are ways to place the first meals. Then for the other two people, there are ways to serve their meals. Thus, there are ways to serve their meals.
Note: This solution is wholly wrong, whoever wrote this solution doesn't even understand the problem.
Video Solution by Richard Rusczyk =
https://artofproblemsolving.com/videos/amc/2012aimei/331
~ dolphin7
Video Solution
https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.