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

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== Problem ==
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Five people are sitting at a round table. Let <math>f\geq 0</math> be the number of people sitting next to at least 1 female and <math>m\geq0</math> be the number of people sitting next to at least one male. The number of possible ordered pairs <math>(f,m)</math> is
 
Five people are sitting at a round table. Let <math>f\geq 0</math> be the number of people sitting next to at least 1 female and <math>m\geq0</math> be the number of people sitting next to at least one male. The number of possible ordered pairs <math>(f,m)</math> is
  
 
<math> \mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }  </math>
 
<math> \mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }  </math>
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== Solution ==
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Suppose there are more men than women; then there are between zero and two women.
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If there are no women, the pair is <math>(0,5)</math>. If there is one woman, the pair is <math>(2,5)</math>.
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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>(3,5)</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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== See also ==
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{{AHSME box|year=1989|num-b=23|num-a=25}} 
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[[Category: Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:20, 4 February 2016

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 $(3,5)$.

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

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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