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>. | ||
− | 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>\boxed{\rm{(B)}}</math>. |
== Solution 2 == | == 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>. | + | 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 12:27, 19 June 2021
Contents
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 .
Solution 2
Denote as the number of such pairs for
people. Then for
, 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
. Now if we note that
, we have that
, so that the answer is
.
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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