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AIME 2012 I · 第 3 题

AIME 2012 I — Problem 3

专题
Contest Math
难度
L4
来源
AIME

题目详情

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.

解析

Solution 1

Call a beef meal B,B, a chicken meal C,C, and a fish meal F.F. Now say the nine people order meals BBBCCCFFF\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 99 to account for the 99 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 BBCCCFFF\text{BBCCCFFF} to orders BBCCCFFF\text{BBCCCFFF} with no matches. The two people who ordered BB's can either both get CC's, both get FF's, or get one CC and one F.F. We proceed with casework.

  • If the two BB people both get CC's, then the three FF meals left to distribute must all go to the CC people. The FF people then get BBCBBC in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the FF meals to the CC people, and there is only 1 way to order this, as all three meals are the same.
  • If the two BB people both get FF's, the situation is identical to the above and three possibilities arise.
  • If the two BB people get CFCF in some order, then the CC people must get FFBFFB and the FF people must get CCB.CCB. This gives 2⋅3⋅3=182 \cdot 3 \cdot 3 = 18 possibilities.

Summing across the cases we see there are 2424 possibilities, so the answer is 9⋅24=216.9 \cdot 24 = \boxed{216.}

Solution 2- ONLY A COINCIDENCE

We only need to figure out the number of ways to order the string BBBCCCFFFBBBCCCFFF, where exactly one BB is in the first three positions, one CC is in the 4th4^{th} to 6th6^{th} positions, and one FF is in the last three positions. THIS IS AN INCORRECT UNDERSTANDING OF THE PROBLEM!!! IT FAILS FOR OTHER NUMBERS\textbf{THIS IS AN INCORRECT UNDERSTANDING OF THE PROBLEM!!! IT FAILS FOR OTHER NUMBERS}There are 33=273^3=27 ways to place the first 33 meals. Then for the other two people, there are 22 ways to serve their meals. Thus, there are (3⋅2)3=216(3\cdot2)^3=\boxed{216} ways to serve their meals.

Note: This solution gets the correct answer through coincidence and should not be used.\textbf{Note: This solution gets the correct answer through coincidence and should not be used.}

Solution 3

First, choose the person that gets the correct meal. There's obviously 99 ways to do that. Then, we can casework on who gets what type of meal. WLOG the person who had the correct meal had the chicken meal. The remaining chicken meals can be distributed to either one person who ordered beef and another who ordered fish, or can be distributed to two people who ordered the same type of meal that is not chicken. For the first case, it's apparent that there are 99 ways of choosing who gets the two chicken meal, and two ways to order the remaining meals. For the second case, there are 66 ways to choose who gets the two chicken meals and only 11 way of ordering the remaining meals. Therefore, the answer is 9⋅(9⋅2+6⋅1)=2169 \cdot (9 \cdot 2 + 6 \cdot 1) = \boxed {216}.

~Arcticturn

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/331

~ dolphin7

Video Solution

https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S