SOLUTION: 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? Thanks

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Question 967697: 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?
Thanks!
Found 2 solutions by mathswizard, ikleyn:
Answer by mathswizard(1)   (Show Source): You can put this solution on YOUR website!
hmmidk
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.
The total number of placing 4 persons in a row of 8 chairs is   = 8*7*6*5 = 1680.

    Here  represents the number of chosing chairs and 4! represents the number of all possible permutations 
    of 4 person inside the given claster of 4 chairs.


Now, there are 2 major configurations of favorable placings, a) and b)


  a)                            b) 
                     A  B                           B  A
   1  2  3  4  5  6  7  8         1  2  3  4  5  6  7  8


where April and Bobby occupy chairs ## 7 and 8.


For configuration a), we have   = 6*7 = 42 possible placings of two other persons.

For configuration b), we have   = 6*7 = 42 possible placings two other persons  (the same number as in a).


So the number of favorable placings is  42+42 = 84.


Now the probability under the question is  P =  = 0.05.    ANSWER

Solved.


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