Difference between revisions of "2021 WSMO Team Round Problems/Problem 8"
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==Solution== | ==Solution== | ||
| + | First, we deal with Euler and Hypatia. Out of the <math>\binom{10}{2}</math> possible seatings for the two, there are only <math>\frac{10\cdot7}{2}</math> that are valid. Now, given Euler and Hypatia aren't together, we'll calculate the chance Isaac and Gottfried don't sit together. After Euler and Hypatia are seated, note that there are <math>\binom{8}{2}</math> possible seatings of Isaac and Gottfried. Excluding Euler and Hypatia, there are <math>8\cdot5</math> possible arragements of Isaac and Gottfried that are valid, since there are always 2 empty neighboring seats next to Isaac that Gottfried cant sit in. However, over all <math>8</math> possible seatings are Isaac, there are <math>4</math> neighboring seats that are already occupied. Therefore, including Euclid and Hypatia, there are <math>8\cdot5+4 = 44</math> possible valid arragements, where ordering matters. Therefore, our answer is <cmath>\left(\frac{10!}{10}\right)\left(\frac{\frac{10\cdot7}{2}}{\binom{10}{2}}\right)\left(\frac{\frac{44}{2}}{\binom{8}{2}}\right)=51\cdot7!\implies 51+7 = \boxed{56}.</cmath> | ||
| + | ~pinkpig | ||
Latest revision as of 14:14, 9 September 2025
Problem
Isaac, Gottfried, Carl, Euclid, Albert, Srinivasa, René, Adihaya, and Euler sit around a round table (not necessarily in that order). Then, Hypatia takes a seat. There are 
possible seatings where Euler doesn't sit next to Hypatia and Isaac doesn't sit next to Gottfried, where 
is maximized. Find 
. (Rotations are not distinct, but reflections are).
Proposed by mahaler
Solution
First, we deal with Euler and Hypatia. Out of the 
possible seatings for the two, there are only 
that are valid. Now, given Euler and Hypatia aren't together, we'll calculate the chance Isaac and Gottfried don't sit together. After Euler and Hypatia are seated, note that there are 
possible seatings of Isaac and Gottfried. Excluding Euler and Hypatia, there are 
possible arragements of Isaac and Gottfried that are valid, since there are always 2 empty neighboring seats next to Isaac that Gottfried cant sit in. However, over all 
possible seatings are Isaac, there are 
neighboring seats that are already occupied. Therefore, including Euclid and Hypatia, there are 
possible valid arragements, where ordering matters. Therefore, our answer is ![\[\left(\frac{10!}{10}\right)\left(\frac{\frac{10\cdot7}{2}}{\binom{10}{2}}\right)\left(\frac{\frac{44}{2}}{\binom{8}{2}}\right)=51\cdot7!\implies 51+7 = \boxed{56}.\]](http://latex.artofproblemsolving.com/c/e/e/ceef5ac85d69cf6e9852e6ecec2d9d5d21d8b8ff.png)
~pinkpig
