Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other? Choose the correct choice from the following choices:
A) 1/30
B) 1/15
C) 1/5
D) 2/5
E) 7/20
Total no of empty space is 5 so total no of ways A B C D and E can sit is 5 !   Solution 1:   Now condition is C&amp;E will sit together   If they sit at 1st and 2nd place . # of ways will be 1*3! but C&amp;E can be either at 1 or at 2 so # of ways will be 1*3!*2!   Similarly they can sit together at position 2 and 3 or 3 and 4 or 4 and 5   So total # ways will be 1*3!*2! + 1*3!*2! + 1*3!*2! + 1*3!*2! = 4 *1*3! *2 !   So probability is (4! * 2! ) / 5! = 2/5   Solution 2:   Consider C*E as 1 person . So we have 4 ppl and 4 places . # of ways they can sit is 4!   C&amp;E can sit as C&amp;E or E&amp;C so total # of ways is 4! * 2!   So probability will be 2/5   Answer:2/5
<result>D</result>
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?