2012 AIME I Problems/Problem 3
Problem 3
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
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.
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.
- 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
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 |