: In a certain community, there are 3 families consisting of a single parent and 1 child, 3 families consisting of a single parent and 2 children, 5 families consisting of 2 parents and a single child, 7 families consisting of 2 parents and 2 children, and 6 families consisting of 2 parents and 3 children. If a parent and child from the same family are to be chosen, how many possible choices are there?

Question
Answer:
To find the total number of possible choices of a parent and child from the same family, we'll calculate the choices for each type of family and sum them up. Let's denote: - A: Single parent with 1 child families - B: Single parent with 2 children families - C: Two parents with 1 child families - D: Two parents with 2 children families - E: Two parents with 3 children families For each type of family, the number of possible choices is the product of the number of families and the number of choices within each family. 1. Single parent with 1 child families (A): Number of choices in each family: 1 (the single parent and the child) Total choices for A: $$\(3 \times 1 = 3\)$$ 2. Single parent with 2 children families (B): Number of choices in each family: 2 (the single parent with each of the two children) Total choices for B: $$\(3 \times 2 = 6\)$$ 3. Two parents with 1 child families (C): Number of choices in each family: 2 (either of the two parents with the child) Total choices for C: $$\(5 \times 2 = 10\)$$ 4. Two parents with 2 children families (D): Number of choices in each family: 3 (either of the two parents with each of the two children) Total choices for D: $$\(7 \times 3 = 21\)$$ 5. Two parents with 3 children families (E): Number of choices in each family: 4 (either of the two parents with each of the three children) Total choices for E: $$\(6 \times 4 = 24\)$$ Now, sum up the choices for each type of family to get the total number of possible choices: $$\[ \text{Total choices} = A + B + C + D + E = 3 + 6 + 10 + 21 + 24 = 64 \]$$ So, there are a total of 64 possible choices of a parent and child from the same family.
solved
general 11 months ago 1695