Difference between revisions of "2012 AIME I Problems/Problem 3"
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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. | 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== | + | ==Solution 1== |
Call a beef meal <math>B,</math> a chicken meal <math>C,</math> and a fish meal <math>F.</math> Now say the nine people order meals <math>\text{BBBCCCFFF}</math> 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 <math>9</math> to account for the <math>9</math> 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. | Call a beef meal <math>B,</math> a chicken meal <math>C,</math> and a fish meal <math>F.</math> Now say the nine people order meals <math>\text{BBBCCCFFF}</math> 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 <math>9</math> to account for the <math>9</math> 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. | ||
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Summing across the cases we see there are <math>24</math> possibilities, so the answer is <math>9 \cdot 24 = \boxed{216.}</math> | Summing across the cases we see there are <math>24</math> possibilities, so the answer is <math>9 \cdot 24 = \boxed{216.}</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | Since there are three people who want each type of meal there are <math>3!=6</math> ways the waiter can serve one type of dish. For the waiter to serve all the dishes, there are <math>6\times 6 \times 6 = \boxed{216}</math> total ways to serve all the corresponding dishes to the corresponding people. | ||
| + | ~MathleteMA | ||
== See also == | == See also == | ||
Revision as of 05:33, 25 February 2017
Contents
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 
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.
in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the 
in some order, then the 
and the 
This gives 
possibilities. Summing across the cases we see there are 
possibilities, so the answer is 
Solution 2
Since there are three people who want each type of meal there are 
ways the waiter can serve one type of dish. For the waiter to serve all the dishes, there are 
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. 
