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 "1989 AHSME Problems/Problem 24"

m (corrected number in one pair)
(→‎Solution)
Line 13: Line 13:
  
 
All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so <math>\rm{(B)}</math>.
 
All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so <math>\rm{(B)}</math>.
 +
 +
== Solution 2 ==
 +
Denote <math>T_n</math> as the number of such pairs for <math>n</math> people. Then for <math>T_{n-1}</math>, when we add an extra spot, we can either have a male or female giving two options. Note that these two options however double the value of <math>T_{n-1}</math>. Now if we note that <math>T_2=1</math>, we have that <math>T_5=8</math>, so that the answer is <math>\rm{(B)8}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:26, 19 June 2021

Problem

Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is

$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }$

Solution

Suppose there are more men than women; then there are between zero and two women.

If there are no women, the pair is $(0,5)$. If there is one woman, the pair is $(2,5)$.

If there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $(4,5)$ and $(3,4)$.

All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so $\rm{(B)}$.

Solution 2

Denote $T_n$ as the number of such pairs for $n$ people. Then for $T_{n-1}$, when we add an extra spot, we can either have a male or female giving two options. Note that these two options however double the value of $T_{n-1}$. Now if we note that $T_2=1$, we have that $T_5=8$, so that the answer is $\rm{(B)8}$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_24&oldid=156374"