Difference between revisions of "1989 AHSME Problems/Problem 30"
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Suppose that the class tried every configuration. Boy <math>i</math> and girl <math>j</math> would stand next to each other in <math>2</math> different orders, in <math>19</math> different positions, <math>18!</math> times each. Summing over all <math>i,j</math> gives <math>7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!</math>, so the average value of <math>S</math> is <math>\boxed{9\tfrac1{10}(A)}</math>. | Suppose that the class tried every configuration. Boy <math>i</math> and girl <math>j</math> would stand next to each other in <math>2</math> different orders, in <math>19</math> different positions, <math>18!</math> times each. Summing over all <math>i,j</math> gives <math>7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!</math>, so the average value of <math>S</math> is <math>\boxed{9\tfrac1{10}(A)}</math>. | ||
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== See also == | == See also == | ||
Revision as of 20:29, 22 January 2016
Problem
Suppose that 7 boys and 13 girls line up in a row. Let 
be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row 
we have that 
. The average value of (if all possible orders of these 20 people are considered) is closest to
Solution
Suppose that the class tried every configuration. Boy 
and girl 
would stand next to each other in 
different orders, in 
different positions, 
times each. Summing over all 
gives 
, so the average value of is 
.
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last Question | |
| 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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