3 persons (1 couple and 1 single) are seated at random in a row of 5 chairs. What is the probability that the couple does not sit together? Pick one
A) 5/7
B) 4/5
C) 2/5
D) 3/5
E) 11/8
Let's find the probability that a couple sits together (right next to each other) and subtract that value from 1.   Total # of ways 3 persons C1C1, C2C2 and SS to be seated in a row of 5 seats is 5!2!=605!2!=60. Consider this, we are interested in arrangement of C1, C2, S, E, EC1, C2, S, E, E, so in arrangement of 5 letters out of which 2 E's are identical (E denotes an empty seat);   # of ways for a couple to sit together is 4!2!∗2=244!2!∗2=24. Consider a couple as a single unit: {C1,C2}, S, E, E{C1,C2}, S, E, E, so we have total of 4 units out of which 2 E's are identical, # of arrangement of these units is 4!2!4!2!, but C1C1, C2C2 within their unit can be arranged in 2 ways ({C1,C2}{C1,C2} or {C2,C1}{C2,C1}), so total # of arrangement for this case is 4!2!∗2=244!2!∗2=24;   P=1−2460=35P=1−2460=35.   Answer: 3/5.
<result>D</result>
3 persons (1 couple and 1 single) are seated at random in a row of 5 chairs. What is the probability that the couple does not sit together?