Difference between revisions of "1989 AHSME Problems/Problem 24"
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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, | + | 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>. | ||
Revision as of 14:14, 7 March 2021
Problem
Five people are sitting at a round table. Let 
be the number of people sitting next to at least 1 female and 
be the number of people sitting next to at least one male. The number of possible ordered pairs 
is
Solution
Suppose there are more men than women; then there are between zero and two women.
If there are no women, the pair is 
. If there is one woman, the pair is 
.
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 
and 
.
All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so 
.
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. 
