Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?  Options: A) 1/5 B) 1/4 C) 3/8 D) 2/5 E) 1/2
Let A be any person from Couple A and B be any person from Couple B.   There are only two cases how the single person could be distributed to split the couples:   (1) * A1 * B1 * A2 * B2 *   (2) A1 B1 * A2 B2   where * denotes the possible seating arrangements for the single person.   (1) There are 5∗2!∗2!∗2!=40 different seating arrangements. We multiply by 2! because there are two options for couple A and two options for couple B to be seated. Furthermore, couple B could be seated to the left of A. Consider these possibilities:   * A2 * B1 * A1 * B2 *   * A1 * B2 * A3 * B1 *   * B1 * A1 * B2 * A2 *   etc.   (2) There are 2!∗2!∗2!=8 different seating arrangements. Same reason as above. Consider these possibilities:   A2 B1 * A1 B2   B1 A1 * B2 A2   etc.   Overall we have  40+8= 
<gadget id="calculator"> 40+8</gadget>
<output>48</output>
48 possible arrangements and 5!=120 ways to arrange 5 people. The probability is 48/120=2/5   Answer: 2/5.
<result>D</result>
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?