Question 1207172
.
April, Bill , Candace, and Bobby are to be seated at random in a row of
8 chairs. What is the probability that April and Bobby will occupy 
the seats at the end of the row?
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<pre>
April   can occupy any of 8 seats;
Bill    can occupy any of 7 remaining seats;
Candace can occupy any of 6 remaining seats;
Bobby   can occupy any of 5 remaining seats.


In all, there are 8*7*6*5 = 1680 possible different placements.


The favorable placements are 

    (a)  April is at the most left seat; Bobby is at the most right seat;
         two other persons in any 2 of 6 remaining seats,
         which gives 6*5 = 30 options.


    (b)  April is at the most right seat; Bobby is at the most left seat;
         two other persons in any 2 of 6 remaining seats,
         which gives 6*5 = 30 other options.


Therefore, the desired probability is  P = {{{(30+30)/1680}}} = {{{60/1680}}} = {{{1/28}}}.


<Y>ANSWER</U>.  The desired probability is 1/28.
</pre>

Solved.


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The problem's formulation is not totally clear and admits different interpretations.


My interpretation was that April and Bobby occupy the seats at two opposite ends of the row.


Other possible interpretation can be that April and Bobby occupy two adjacent seats at some of the two ends.


When a problem admits different interpretations, it is a fault of its creator.