There are three teams `x`,`x+1` and `y` childrens and total number of childrens in the teams is `24`. if two childrens of the same team do not fight, then

Options

A. (a) maximum number of fights is 190
B. (b) maximum number of fights is 191
C. (c) maximum number of fights occur when x=7
D. (d) maximum number of fights occur when x=8

Explanation

To solve the problem step by step, we will first define the variables and set up the equations based on the information given.

### Step 1: Define the Variables
Let:
- Team 1 has `x` children.
- Team 2 has `x + 1` children.
- Team 3 has `y` children.

### Step 2: Set Up the Equation
According to the problem, the total number of children in all teams is 24. Therefore, we can write the equation:
\[ x + (x + 1) + y = 24 \]
This simplifies to:
\[ 2x + 1 + y = 24 \]

### Step 3: Solve for y
Rearranging the equation to solve for `y` gives:
\[ y = 24 - 2x - 1 \]
\[ y = 23 - 2x \]

### Step 4: Calculate Total Possible Fights
The total number of ways to choose 2 children from 24 is given by the combination formula:
\[ \binom{24}{2} = \frac{24 \times 23}{2} = 276 \]

Next, we need to subtract the fights that occur within each team since children from the same team do not fight.

- For Team 1 (with `x` children):
\[ \binom{x}{2} = \frac{x(x-1)}{2} \]

- For Team 2 (with `x + 1` children):
\[ \binom{x + 1}{2} = \frac{(x + 1)x}{2} \]

- For Team 3 (with `y` children):
\[ \binom{y}{2} = \frac{(23 - 2x)(22 - 2x)}{2} \]

### Step 5: Set Up the Equation for Total Fights
The total number of fights `N` can be expressed as:
\[ N = \binom{24}{2} - \left( \binom{x}{2} + \binom{x + 1}{2} + \binom{y}{2} \right) \]
Substituting the expressions we have:
\[ N = 276 - \left( \frac{x(x-1)}{2} + \frac{(x + 1)x}{2} + \frac{(23 - 2x)(22 - 2x)}{2} \right) \]

### Step 6: Simplify the Expression for N
Combining the terms:
\[ N = 276 - \left( \frac{x^2 - x + x^2 + x + (23 - 2x)(22 - 2x)}{2} \right) \]
\[ = 276 - \left( \frac{2x^2 + (23 - 2x)(22 - 2x)}{2} \right) \]

### Step 7: Expand and Simplify
Expanding \( (23 - 2x)(22 - 2x) \):
\[ = 506 - 46x + 4x^2 \]
Thus:
\[ N = 276 - \left( \frac{2x^2 + 506 - 46x + 4x^2}{2} \right) \]
\[ = 276 - \left( \frac{6x^2 - 46x + 506}{2} \right) \]
\[ = 276 - 3x^2 + 23x - 253 \]
\[ = 23 + 23x - 3x^2 \]

### Step 8: Maximize N
To find the maximum number of fights, we differentiate `N` with respect to `x` and set the derivative to zero:
\[ \frac{dN}{dx} = 23 - 6x \]
Setting this equal to zero gives:
\[ 23 - 6x = 0 \]
\[ x = \frac{23}{6} \approx 3.83 \]

### Step 9: Check Integer Values
Since `x` must be an integer, we check `x = 3` and `x = 4` to find the maximum fights.

1. For `x = 3`:
- `y = 23 - 2(3) = 17`
- Calculate `N`.

2. For `x = 4`:
- `y = 23 - 2(4) = 15`
- Calculate `N`.

### Step 10: Conclusion
After calculating for both values, we find the maximum number of fights.