Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement.
Arrangement – AIME 2012
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.
- is 107
- is 216
- is 840
- cannot be determined from the given information
Key Concepts
Arrangements
Algebra
Number Theory
Check the Answer
But try the problem first…
Answer: is 216.
AIME, 2012, Question 3
Combinatorics by Brualdi
Try with Hints
First hint
Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions
Second Hint
one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.
Final Step
Then \((3.2)^3={216}\)ways
Other useful links
- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s
Related Program
Math Olympiad Program… taught by Olympians and researchers
