A man wants to distribute 101 coins a rupee each, among his 3 sons with the condition that no one receives more money than the combined total of other two. The number of ways of doing this is :-

Options

A. `.^(103)C_(2) - 3^(52)C_(2)`
B. `(.^(103)C_(2))/(3)`
C. `1275`
D. `(.^(103)C_(2))/(6)`

Answer

To solve the problem of distributing 101 coins among three sons such that no son receives more money than the combined total of the other two, we can follow these steps:

### Step 1: Define the Variables
Let the amounts received by the three sons be \( x \), \( y \), and \( z \). The total amount of money distributed is given by:
\[ x + y + z = 101 \]

### Step 2: Set Up the Conditions
According to the problem, no son can receive more than the combined total of the other two. This gives us the following inequalities:
1. \( x \leq y + z \)
2. \( y \leq x + z \)
3. \( z \leq x + y \)

### Step 3: Simplify the Conditions
From the first inequality \( x \leq y + z \), we can substitute \( y + z \) with \( 101 - x \):
\[ x \leq 101 - x \]
This simplifies to:
\[ 2x \leq 101 \quad \Rightarrow \quad x \leq 50.5 \]
Since \( x \) must be an integer, we have:
\[ x \leq 50 \]
Similarly, we can derive:
\[ y \leq 50 \quad \text{and} \quad z \leq 50 \]

### Step 4: Count the Solutions
Now we need to count the non-negative integer solutions to the equation \( x + y + z = 101 \) under the constraints \( x \leq 50 \), \( y \leq 50 \), and \( z \leq 50 \).

### Step 5: Use the Stars and Bars Method
Without any constraints, the number of non-negative integer solutions to \( x + y + z = 101 \) can be found using the stars and bars method:
\[ \text{Number of solutions} = \binom{101 + 3 - 1}{3 - 1} = \binom{103}{2} \]

### Step 6: Subtract the Invalid Cases
Next, we need to subtract the cases where one variable exceeds 50. Assume \( x > 50 \). Let \( x' = x - 51 \) (where \( x' \geq 0 \)). The equation becomes:
\[ x' + y + z = 50 \]
The number of non-negative integer solutions is:
\[ \binom{50 + 3 - 1}{3 - 1} = \binom{52}{2} \]
The same reasoning applies for \( y > 50 \) and \( z > 50 \), so we subtract this count three times:
\[ 3 \times \binom{52}{2} \]

### Step 7: Add Back the Over-Subtracted Cases
Finally, we need to add back the cases where two variables exceed 50. Assume \( x > 50 \) and \( y > 50 \). Let \( x' = x - 51 \) and \( y' = y - 51 \). The equation becomes:
\[ x' + y' + z = -1 \]
This has no non-negative solutions, so we do not need to add anything back.

### Step 8: Calculate the Final Count
Putting it all together, we have:
\[ \text{Total valid solutions} = \binom{103}{2} - 3 \times \binom{52}{2} \]

Calculating the binomial coefficients:
\[ \binom{103}{2} = \frac{103 \times 102}{2} = 5253 \]
\[ \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \]
Thus,
\[ \text{Total valid solutions} = 5253 - 3 \times 1326 = 5253 - 3978 = 1275 \]

### Final Answer
The number of ways to distribute the coins is \( \boxed{1275} \).