Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 6"
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There are <math>4^5</math> ways for the friends to choose houses. Now, we have friends can meet in pairs, so <math>(2,3)</math> is the only way they can meet. There are <math>20</math> ways for them to choose the <math>2</math> houses to meet, and <math>3</math> ways to rotate the "extra" partner (2 houses are already selected) The answer is <cmath>\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.</cmath> ~Geometry285 | There are <math>4^5</math> ways for the friends to choose houses. Now, we have friends can meet in pairs, so <math>(2,3)</math> is the only way they can meet. There are <math>20</math> ways for them to choose the <math>2</math> houses to meet, and <math>3</math> ways to rotate the "extra" partner (2 houses are already selected) The answer is <cmath>\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.</cmath> ~Geometry285 | ||
| + | == Solution 2 == | ||
| + | <cmath>\frac{2^2 \cdot 3 \cdot 5}{4^{5}} \implies \frac{15}{256}</cmath>Therefore, <math>15+256=\boxed{271}</math> | ||
| + | - kante314 - | ||
==See also== | ==See also== | ||
Latest revision as of 09:01, 12 July 2021
Contents
Problem
Five friends decide to meet together for a party. However, they did not plan the party well, and at noon, every friend leaves their own house and travels to one of the other four friends' houses, chosen uniformly at random. The probability that every friend sees another friend in the house they chose can be expressed in the form 
. If 
and 
are relatively prime positive integers, find 
.
Solution
There are 
ways for the friends to choose houses. Now, we have friends can meet in pairs, so 
is the only way they can meet. There are 
ways for them to choose the 
houses to meet, and 
ways to rotate the "extra" partner (2 houses are already selected) The answer is ![\[\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.\]](http://latex.artofproblemsolving.com/8/3/3/83359b8a3783ced14abddd26d120bc8fd335dd6a.png)
~Geometry285
Solution 2
![\[\frac{2^2 \cdot 3 \cdot 5}{4^{5}} \implies \frac{15}{256}\]](http://latex.artofproblemsolving.com/2/a/5/2a5bf58141adcec9b3efeaa50122760dcf37c627.png)
Therefore, 
- kante314 -
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
