Final answer:
The maximum number of fights that can take place is 33, calculated by establishing the maximum integer value of x such that the total number of children does not exceed 10 and finding all possible fight pairings not involving siblings.
Step-by-step explanation:
Given that Mr. A has x children and Ms. B has x+1 children from previous marriages, and together they have 10 children in total, we can establish that Mr. A and Ms. B together will have 10 - (x + x + 1) = 10 - 2x - 1 = 9 - 2x children of their own. To find the maximum number of fights, we first recognize that fights can only occur between children of different parents, as specified that no fights occur between children of the same parents.
We can create pairs of children who can potentially fight which are: children of Mr. A with children of Ms. B and children of Mr. A or Ms. B with their common children. If Mr. A has x children and Ms. B has x+1, then between them, there could be x * (x + 1) possible fights. The remaining 9 - 2x children can fight with Mr. A's x children and Ms. B's x+1 children, that adds (9 - 2x) * (2x + 1) fights. To find the maximum number of fights we need to maximize x, knowing that all terms must remain positive and integer values, as they represent actual children.
By trial and error, starting with x = 1 and increasing, we find that x = 3 is the largest value before the total surpasses 10 children, which gives 3 children for Mr. A, 4 for Ms. B, and 3 common children. The maximum number of fights is 3 * 4 + 3 * (3+4) = 12 + 21 = 33.