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}}</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>. |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:52, 6 June 2014
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 
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
