John has x children by his first wife. Mary has `(x+1)` children by her first husnand. They marry and hasve children of their own. The whole family has 24 children. Assuming tht two children of the same parents do not fight prove that the maximum possible number of fights that can take place is 191.

Answer

To solve the problem, we will follow a systematic approach to find the maximum number of fights that can occur among the children of John and Mary.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( x \) be the number of children John has from his first wife.
- Mary has \( x + 1 \) children from her first husband.
- Let \( y \) be the number of children John and Mary have together.

2. **Total Children**:
- The total number of children in the family is given as 24. Therefore, we can write the equation:
\[ x + (x + 1) + y = 24 \]
- Simplifying this gives:
\[ 2x + y + 1 = 24 \implies y = 23 - 2x \]

3. **Fights Calculation**:
- We need to calculate the total number of fights \( z \) that can occur among the children. The fights can occur between:
- John's children and Mary's children.
- John's children and their own children.
- Mary's children and their own children.

4. **Fights Between Different Groups**:
- The fights can be calculated as follows:
- Fights between John's children (\( x \)) and Mary's children (\( x + 1 \)):
\[ z_1 = x \cdot (x + 1) \]
- Fights between John's children (\( x \)) and their own children (\( y \)):
\[ z_2 = x \cdot y \]
- Fights between Mary's children (\( x + 1 \)) and their own children (\( y \)):
\[ z_3 = (x + 1) \cdot y \]

5. **Total Fights**:
- Therefore, the total number of fights \( z \) can be expressed as:
\[ z = z_1 + z_2 + z_3 = x(x + 1) + xy + (x + 1)y \]
- Expanding this:
\[ z = x^2 + x + xy + xy + y = x^2 + 2xy + x + y \]

6. **Substituting for \( y \)**:
- Substitute \( y = 23 - 2x \) into the equation:
\[ z = x^2 + 2x(23 - 2x) + x + (23 - 2x) \]
- Simplifying this:
\[ z = x^2 + 46x - 4x^2 + x + 23 - 2x \]
\[ z = -3x^2 + 45x + 23 \]

7. **Finding Maximum Fights**:
- To find the maximum number of fights, we need to determine the maximum value of the quadratic function \( z = -3x^2 + 45x + 23 \).
- The maximum occurs at the vertex of the parabola, given by:
\[ x = -\frac{b}{2a} = -\frac{45}{2 \cdot -3} = \frac{45}{6} = 7.5 \]
- Since \( x \) must be an integer, we evaluate \( z \) at \( x = 7 \) and \( x = 8 \).

8. **Calculating for \( x = 7 \)**:
- If \( x = 7 \):
\[ y = 23 - 2(7) = 9 \]
\[ z = -3(7^2) + 45(7) + 23 = -147 + 315 + 23 = 191 \]

9. **Calculating for \( x = 8 \)**:
- If \( x = 8 \):
\[ y = 23 - 2(8) = 7 \]
\[ z = -3(8^2) + 45(8) + 23 = -192 + 360 + 23 = 191 \]

10. **Conclusion**:
- In both cases, the maximum number of fights \( z \) is 191. Thus, we conclude that the maximum possible number of fights that can take place is:
\[ \boxed{191} \]