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Difference between revisions of "2012 AIME I Problems/Problem 3"

(Created page with "== 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 m...")
 
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== 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.
 
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.
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==Solution==
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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>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.
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The problem we must solve is to distribute meals <math>BBCCCFFF</math> to orders <math>BBCCCFFF</math> with no matches. The two people who ordered <math>B</math>'s can either both get <math>C</math>'s, both get <math>F</math>'s, or get one <math>C</math> and one <math>F.</math> We proceed with casework.
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<UL>
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<LI> If the two <math>B</math> people both get <math>C</math>'s, then the three <math>F</math> meals left to distribute must all go to the <math>C</math> people. The <math>F</math> people then get <math>BBC</math> in some order, which gives three possibilities. </LI>
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<LI> If the two <math>B</math> people both get <math>F</math>'s, the situation is identical to the above and three possibilities arise. </LI>
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<LI> If the two <math>B</math> people get <math>CF</math> in some order, then the <math>C</math> people must get <math>FFB</math> and the <math>F</math> people must get <math>CCB.</math> This gives <math>2 \cdot 3 \cdot 3 = 18</math> possibilities. </LI>
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</UL>
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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>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2012|n=I|num-b=2|num-a=4}}

Revision as of 02:09, 17 March 2012

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 $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $BBBCCCFFF$ 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.

The problem we must solve is to distribute meals $BBCCCFFF$ to orders $BBCCCFFF$ 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.
  • 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 $9 \cdot 24 = \boxed{216.}$

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
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