1. Problem: Find the number of ways to arrange 2 couples in a row under different restrictions. 2. Formula: The number of permutations of $n$ distinct objects is given by $$n!$$. 3. Important rules: - When objects are grouped, treat the group as a single object first, then multiply by the permutations inside the group. - Restrictions reduce the total number of permutations. 4. Part 1a: No restriction. - There are 4 people (2 couples), so total permutations are $$4! = 24$$. 5. Part 1b: Men are grouped together. - Treat the 2 men as one group plus the 2 women, so 3 objects total. - Permutations of these 3 objects: $$3! = 6$$. - Permutations inside the men group: $$2! = 2$$. - Total permutations: $$3! \times 2! = 6 \times 2 = 12$$. 6. Part 1c: Men and women are grouped together. - Treat men as one group and women as another group, so 2 groups total. - Permutations of these 2 groups: $$2! = 2$$. - Permutations inside men group: $$2! = 2$$. - Permutations inside women group: $$2! = 2$$. - Total permutations: $$2! \times 2! \times 2! = 2 \times 2 \times 2 = 8$$. 7. Part 2a: Arrange 2 couples in a row of 3 seats with no restriction. - We have 4 people but only 3 seats, so number of permutations is $$P(4,3) = \frac{4!}{(4-3)!} = \frac{24}{1} = 24$$. 8. Part 2b: The first seat is a man. - Number of men: 2. - Choose a man for the first seat: $$2$$ ways. - Remaining seats: 2 seats, remaining people: 3. - Permutations for remaining seats: $$P(3,2) = \frac{3!}{(3-2)!} = 6$$. - Total permutations: $$2 \times 6 = 12$$. 9. Part 2c: The first two seats are women. - Number of women: 2. - Arrange 2 women in first two seats: $$2! = 2$$ ways. - Remaining seat: 1 seat, remaining people: 2 (both men). - Choose 1 man for last seat: $$2$$ ways. - Total permutations: $$2 \times 2 = 4$$. Final answers: 1a. 24 1b. 12 1c. 8 2a. 24 2b. 12 2c. 4