4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 3 end up sitting in the same chair that they sat in during the first half? Choose the correct answer:
A) 3/24 B) 9/24 C) 15/24 D) 1/8 E) 21/24
Originally seated A B C D   now when after they get up and when they sit back again .   1st- A has option to sit on 3 seats ( apart from his previous seat . thus he now sits on B's seat.)   2nd- Similarly B has option to sit on 3 seats ( because A has already occupied B's previous seat, thus B sits on a's seat.)   3rd- Now C has only 1 option to sit on D's seat . and similarly D also has one option to sit on C's seat.)   hence total favourable outcomes   3*1*1= 
<gadget id="calculator">  3*1*1</gadget>
<output>3</output>
3   and total possible outcomes=4!=24   probability of the favourable outcome= 1/8. 1/8
<result>D</result>
4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 3 end up sitting in the same chair that they sat in during the first half?