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Difference between revisions of "1989 AHSME Problems/Problem 24"

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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 <math>(4,5)</math> and <math>(3,4)</math>.
 
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 <math>(4,5)</math> and <math>(3,4)</math>.
  
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All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so <math>\rm{(B)}</math>.
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All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so <math>\boxed{\rm{(B)}}</math>.
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== Solution 2 ==
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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>\boxed{\rm{(B)8}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 11:27, 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 $\boxed{\rm{(B)}}$.

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 $\boxed{\rm{(B)8}}$.

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

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