If 5 married couples are arranged in a row, find the probability that no husband sits next to his wife? (Hint: Use the inclusion-exclusion principle.)

Answer 1

1 - 9*1/9+C(8,2)*1/9*1/7-C(7,3)*1/9*1/7*1/5+C(6,4)*1/9*1/7*1/5*1/3-C(5,5)*1/9*1/7*1/5*1/3*1/1 = 47/135 =.346148

You remove the 1 pair, add the two pairs, subtract the three pairs, add the four pairs, and remove the 5 pairs.

To explain the formula above:

To get the number for each number of pairs, you find the number of ways to get that number of pairs * the probability of these being pairs.

For example, for two pairs, in the first pair, the probability of the second person being the spouse of the first person is 1/9 (1 out of 9 remaining people is the spouse); then, there are 8 people left, so once someone is selected in the first position, there are seven people left, one of which is the spouse, so the probability is 1/7. Since each pair takes two positions rather than one, there are C(8,2) possible positions for the pairs.

Answer 2

1 - 9*1/9+C(8,2)*1/9*1/7-C(7,3)*1/9*1/7*1/5+C(6,4)*1/9*1/7*1/5*1/3-C(5,5)*1/9*1/7*1/5*1/3*1/1 = 47/135 =.346148

You remove the 1 pair, add the two pairs, subtract the three pairs, add the four pairs, and remove the 5 pairs.

To explain the formula above:

To get the number for each number of pairs, you find the number of ways to get that number of pairs * the probability of these being pairs.

For example, for two pairs, in the first pair, the probability of the second person being the spouse of the first person is 1/9 (1 out of 9 remaining people is the spouse); then, there are 8 people left, so once someone is selected in the first position, there are seven people left, one of which is the spouse, so the probability is 1/7. Since each pair takes two positions rather than one, there are C(8,2) possible positions for the pairs.