Seven couples attended Shauna’s Online Couple Counseling for a group therapy session during a pandemic. To facilitate a role-playing exercise, Shauna selects one of her 14 clients to demonstrate the activity. She then instructs the remaining attendees to pair up randomly with anyone other than their own partner, leaving one person excluded from the exercise. If Shauna selects Paul to assist her in demonstrating the activity and then chooses one of the randomly created pairs to role-play with her and Paul, select for Possible pairs the total number of different pairs that can role-play with Shauna and Paul, and select for Pairs with Paul's partner the total number of different pairs that include Paul's original partner and can role-play with Shauna and Paul. Make only two selections, one in each column.

Official Solution: Without Paul, 13 people remain to form pairs. Shauna and Paul can face any two people from these 13, except for 6 pairs that are actual couples among them. Therefore, the total number of different pairs that can role-play with Shauna and Paul is C^2_{13} - 6 = 72 . As for Paul's original partner, that partner can pair with any of the other 12 people, resulting in a total of 12 different pairs that include Paul's original partner. Correct answer: Possible pairs "72" Pairs with Paul's partner "12" GMAT-Club-Forum-zte7xxkp.png

1. Let's split everyone into their original groups and take away Shauna with Paul: Shauna ( Paul , Paul's Partner) ( a_1 , a_2 ) ( b_1 , b_2 ) ( c_1 , c_2 ) ( x_1 , x_2 ) ( y_1 , y_2 ) ( z_1 , z_2 ) 2. Find the total number of different pairs that can role-play with Shauna and Paul. There are two cases, depending whether we pick Paul's Partner. - The pair includes Paul's Partner. This is only \binom{1}{1} cases. Then, for the other person we can pick anyone else since they won't be partners with Paul's Partner. So, \binom{12}{1} cases and altogether there would be \binom{1}{1} * \binom{12}{1} = 1 * 12 = 12 different pairs. - The pair doesn't include Paul's Partner. There are no restrictions, so, there are \binom{12}{1} cases to pick for the first person in the pair. However, picking the second person is from everyone except the partner of the person who was first picked. In other words, we can't pick b_1 and b_2 . This has \binom{10}{1} cases and together there are \binom{12}{1} * \binom{10}{1} = 12 * 10 = 120 pairs. These pairs are double counting, for example, we can pick x_1 and c_2 or c_2 and x_1 - which are the same. So, there are actually \frac{120}{2} = 60 pairs. Now, we need to sum up these two cases to get 12 + 60 = 72 possible pairs. 3. Find the total number of different pairs that include Paul's original partner and can role-play with Shauna and Paul. We calculated this earlier and got 12 pairs. 4. Our answer will be: Possible pairs - 72 and Pairs with Paul's partner - 12 .