Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2012 AIME I Problems/Problem 3

Revision as of 05:33, 25 February 2017 by Mathletema (talk | contribs)

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 1

Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\text{BBBCCCFFF}$ 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 $9$ to account for the $9$ 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 $\text{BBCCCFFF}$ to orders $\text{BBCCCFFF}$ with no matches. The two people who ordered $B$'s can either both get $C$'s, both get $F$'s, or get one $C$ and one $F.$ We proceed with casework.

  • If the two $B$ people both get $C$'s, then the three $F$ meals left to distribute must all go to the $C$ people. The $F$ people then get $BBC$ in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the $F$ meals to the $C$ people, and there is only 1 way to order this, as all three meals are the same.
  • If the two $B$ people both get $F$'s, the situation is identical to the above and three possibilities arise.
  • If the two $B$ people get $CF$ in some order, then the $C$ people must get $FFB$ and the $F$ people must get $CCB.$ This gives $2 \cdot 3 \cdot 3 = 18$ possibilities.

Summing across the cases we see there are $24$ possibilities, so the answer is $9 \cdot 24 = \boxed{216.}$

Solution 2

Since there are three people who want each type of meal there are $3!=6$ ways the waiter can serve one type of dish. For the waiter to serve all the dishes, there are $6\times 6 \times 6 = \boxed{216}$ total ways to serve all the corresponding dishes to the corresponding people. ~MathleteMA

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_3&oldid=84197"