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Difference between revisions of "2021 WSMO Speed Round Problems/Problem 8"

(Created page with "==Problem== Let <math>n</math> be the number of ways to seat <math>12</math> distinguishable people around a regular hexagon such that rotations do not matter (but reflections...")
 
(→‎Solution)
 
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==Solution==
 
==Solution==
 
Note that there are <math>12!</math> ways to seat the people. Since rotations are considered the same, we have to divide by 6. In addition, since the order in which the two people are seated on each side does not matter, we have to divide by <math>2^6=64.</math> Thus, given the conditions, there are <math>\frac{12!}{64\cdot6}=2^3\cdot3^4\cdot5^2\cdot7^1\cdot11^1.</math> In conclusion, the answer is <math>(3+1)(4+1)(2+1)(1+1)(1+1)=\boxed{240}.</math>
 
Note that there are <math>12!</math> ways to seat the people. Since rotations are considered the same, we have to divide by 6. In addition, since the order in which the two people are seated on each side does not matter, we have to divide by <math>2^6=64.</math> Thus, given the conditions, there are <math>\frac{12!}{64\cdot6}=2^3\cdot3^4\cdot5^2\cdot7^1\cdot11^1.</math> In conclusion, the answer is <math>(3+1)(4+1)(2+1)(1+1)(1+1)=\boxed{240}.</math>
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~pinkpig

Latest revision as of 12:06, 23 December 2021

Problem

Let $n$ be the number of ways to seat $12$ distinguishable people around a regular hexagon such that rotations do not matter (but reflections do), and two people are seated on each side (the order in which they are seated matters). Find the number of divisors of $n$.

Solution

Note that there are $12!$ ways to seat the people. Since rotations are considered the same, we have to divide by 6. In addition, since the order in which the two people are seated on each side does not matter, we have to divide by $2^6=64.$ Thus, given the conditions, there are $\frac{12!}{64\cdot6}=2^3\cdot3^4\cdot5^2\cdot7^1\cdot11^1.$ In conclusion, the answer is $(3+1)(4+1)(2+1)(1+1)(1+1)=\boxed{240}.$

~pinkpig

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