A group consisting of five married couples needs to be divided into two groups. The first group should consist of six people, including at least two married couples. In how many ways can this division be carried out?
To have at least two married couples among the six people means that either a) there are exactly two married couples in the group, or b) the group consists of three married couples. Accordingly, we will separately determine the number of combinations for each case, and the final result will be the sum of these two numbers.

a) If the group of six contains exactly two married couples, let's first calculate how many ways we can choose these two couples. For each married couple, we can choose four other couples, which means we have a total of $5 \cdot 4=20$ possibilities. However, this counts each possibility twice (once with the first couple paired with the second, and once with the second couple paired with the first), so we can choose two couples from the five in $20 / 2=10$ ways.

In the group of six, we still need to add two more individuals from the remaining six, but they cannot be married to each other. Thus, for each of the six individuals, we can choose 4 others, and since each pair is counted twice, this results in $6 \cdot 4 / 2=12$ cases.

Therefore, in this case, the group of six can be formed in $10 \cdot 12=120$ ways.

b) If the group consists of three married couples, we can obtain these three couples in as many ways as we can leave out two couples from the five. We have already calculated this to be 10 ways, so the group of three married couples can also be formed in 10 ways.

In total, therefore, the six people can be selected in $120+10=130$ ways that meet the conditions.