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

m (corrected number in one pair)
 
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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>.
 
If there are no women, the pair is <math>(0,5)</math>. If there is one woman, the pair is <math>(2,5)</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>(3,5)</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>.
  
 
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>.

Latest revision as of 14:14, 7 March 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 $\rm{(B)}$.

See also

1989 AHSME (ProblemsAnswer KeyResources)
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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