Mr. john has x children by his first wife and Ms. Bashu has `x+1` children by her first husband. They marry and have children of their own. The whole family has 10 children. Assuming that two children of the same parents do not fight, Find the maximum number of fight that can take place among children.

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( x \) be the number of children Mr. John has from his first marriage.
- Therefore, Ms. Bashu has \( x + 1 \) children from her first marriage.

2. **Set up the Equation**:
- After they marry, they have their own children as well. Let the number of children they have together be \( y \).
- The total number of children in the family is given as 10. Thus, we can write the equation:
\[ x + (x + 1) + y = 10 \]
- Simplifying this gives:
\[ 2x + 1 + y = 10 \]
- Rearranging, we find:
\[ 2x + y = 9 \]

3. **Express \( y \) in terms of \( x \)**:
- From the equation \( 2x + y = 9 \), we can express \( y \) as:
\[ y = 9 - 2x \]

4. **Find Possible Values for \( x \)**:
- Since \( y \) (the number of children they have together) must be a non-negative integer, we need \( 9 - 2x \geq 0 \). This gives:
\[ 9 \geq 2x \quad \Rightarrow \quad x \leq 4.5 \]
- Since \( x \) must be a whole number, the possible values for \( x \) are \( 0, 1, 2, 3, \) or \( 4 \).

5. **Calculate the Number of Children**:
- If \( x = 3 \):
- Mr. John has 3 children.
- Ms. Bashu has \( 3 + 1 = 4 \) children.
- Together, they have \( y = 9 - 2(3) = 3 \) children.
- Thus, the family consists of:
- Mr. John: 3 children
- Ms. Bashu: 4 children
- Together: 3 children
- Total: \( 3 + 4 + 3 = 10 \) children (which matches the problem statement).

6. **Calculate Maximum Number of Fights**:
- Fights can occur between children from different parents. We will consider fights among:
- Children of Mr. John (3 children).
- Children of Ms. Bashu (4 children).
- Children of both parents (3 children).
- The possible combinations for fights are:
- Choosing 1 child from Mr. John and 1 child from Ms. Bashu: \( 3 \times 4 = 12 \).
- Choosing 1 child from Mr. John and 1 child from their own children: \( 3 \times 3 = 9 \).
- Choosing 1 child from Ms. Bashu and 1 child from their own children: \( 4 \times 3 = 12 \).
- Therefore, the total number of fights is:
\[ 12 + 9 + 12 = 33 \]

### Final Answer:
The maximum number of fights that can take place among the children is **33**.