Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 6"
 
(3 intermediate revisions by 2 users not shown)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
asdf
+
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==
−
#[[2021 JMPSC Invitational Problems|Other 2021 JMPSC Invitational Problems]]
+
#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
−
#[[2021 JMPSC Invitational Answer Key|2021 JMPSC Invitational Answer Key]]
+
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
{{JMPSC Notice}}
 
{{JMPSC Notice}}

Latest revision as of 09:01, 12 July 2021

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 $\frac{m}{n}$. If $m$ and $n$ are relatively prime positive integers, find $m+n$.

Solution

There are $4^5$ ways for the friends to choose houses. Now, we have friends can meet in pairs, so $(2,3)$ is the only way they can meet. There are $20$ ways for them to choose the $2$ houses to meet, and $3$ ways to rotate the "extra" partner (2 houses are already selected) The answer is \[\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.\] ~Geometry285

Solution 2

\[\frac{2^2 \cdot 3 \cdot 5}{4^{5}} \implies \frac{15}{256}\]Therefore, $15+256=\boxed{271}$

- kante314 -

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_6&oldid=158207"