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Difference between revisions of "2019 AIME II Problems/Problem 5"

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Four ambassadors and one advisor for each of them are to be seated at a round table with <math>12</math> chairs numbered in order <math>1</math> to <math>12</math>. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are <math>N</math> ways for the <math>8</math> people to be seated at the table under these conditions. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Four ambassadors and one advisor for each of them are to be seated at a round table with <math>12</math> chairs numbered in order <math>1</math> to <math>12</math>. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are <math>N</math> ways for the <math>8</math> people to be seated at the table under these conditions. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
==Solution==
 
==Solution==
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(NEED SOMEBODY TO CHANGE IT INTO LATEX VERSION)
+
There are <math>4</math> ambassadors and there are <math>6</math> seats for them.
−
There are 4 ambassadors and there are 6 seats for them.
 
 
So we consider the position of the blank seats.
 
So we consider the position of the blank seats.
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There are 15 kinds of versions:
+
There are <math>15</math> kinds of versions:
−
If the two seats are adjacent to each other, there are 6 options, and the ambassadors are sitting in four adjacent seats,and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are 5 kinds of solutions for the advisors to sit. And that’s a 6*5,if we don’t consider the order of the ambassadors.
+
If the two seats are adjacent to each other, there are <math>6</math> options, and the ambassadors are sitting in four adjacent seats, and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are <math>5</math> kinds of solutions for the advisors to sit. And that’s a <math>6\cdot5</math> if we don’t consider the order of the ambassadors.
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We can also get that if the blank seats are opposite, it will be 3*9, if they arenot adjacent and not opposite, it will be 6*8.
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We can also get that if the blank seats are opposite, it will be <math>3\cdot9</math>, if they are not adjacent and not opposite, it will be <math>6\cdot8</math>.
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So the total is 24*(6*3+6*8+3*9)=2520
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So the total is <math>24\cdot(6\cdot3+6\cdot8+3\cdot9)=2520</math>
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And the remainder is 520
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And the remainder is <math>\boxed{520}</math>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2019|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:47, 22 March 2019

Problem

Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$.

Solution

There are $4$ ambassadors and there are $6$ seats for them. So we consider the position of the blank seats. There are $15$ kinds of versions: If the two seats are adjacent to each other, there are $6$ options, and the ambassadors are sitting in four adjacent seats, and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are $5$ kinds of solutions for the advisors to sit. And that’s a $6\cdot5$ if we don’t consider the order of the ambassadors. We can also get that if the blank seats are opposite, it will be $3\cdot9$, if they are not adjacent and not opposite, it will be $6\cdot8$. So the total is $24\cdot(6\cdot3+6\cdot8+3\cdot9)=2520$ And the remainder is $\boxed{520}$

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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