[SOLVED] Probabilities Involving 4 Married Couples 1. The problem statement, all variables and given/known data If 4 married couples are arranged in a row, find the probability that no husband sits next to his wife. 2. The attempt at a solution I will call a married couple a pair and denote n_i as the number of arrangements, given i pairs, where no husband sits next to his wife. There are two ways that I've thought of for counting n_i: (a) n_1 is obviously 0. For i = 2, there are 4 ways to pick the first person, 2 ways to pick the second person and 1 way to pick the third and last persons so n_2 = 4 * 2. Now given another pair, I can stick the husband in front of the first person or behind the first, second, third or fourth person. That's 5 choices. The wife can go in any of the remaining 4 choices. Thus, n_3 = 5 * 4 * n_2 = 5 * 4 * 4 * 2. Given yet another pair, using the same reasoning as before n_4 = 7 * 6 * n_3 = 7 * 6 * 5 * 4 * 4 * 2. The only problem with this method, I think, is that it may not take into account that fact that the pairs are distinct. (b) For i = 4, there are 8 ways to pick the first person. Excluding the spouse of the first person, that leaves 3 pairs of which there are n_3 arrangements. In any of these arrangements, the spouse can behind the first, second, third, ..., or last person, i.e. there are 6 choices for where the spouse could go. Appending one of these arrangements (with the spouse) to the first person produces an arrangement involving 4 pairs. Thus, n_4 = 8 * 6 * n_3. Similarly, n_3 = 6 * 4 * n_2. From (a), n_2 = 4 * 2. Thus, n_4 = 8 * 6 * 6 * 4 * 4 * 2. This method looks more promising. I don't see anything wrong with it. The probability is n_4 / 8!. In both cases, I get the wrong answer. The right answer according to the book is 12/35.