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 Variables
Let the amounts received by the three sons be \(X\), \(Y\), and \(Z\). We have the equation:
\[ X + Y + Z = 101 \]

### Step 2: Set Up the Condition
According to the problem, no son can receive more than the combined total of the other two. This leads to the following inequalities:
\[ X \leq Y + Z \]
\[ Y \leq Z + X \]
\[ Z \leq X + Y \]

### Step 3: Simplify the Conditions
From the first inequality, we can rearrange it as:
\[ X \leq 101 - X \quad \Rightarrow \quad 2X \leq 101 \quad \Rightarrow \quad X \leq 50.5 \]
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: Use Non-Negative Integer Solutions
Now, we need to find the number of non-negative integer solutions to the equation \(X + Y + Z = 101\) under the constraints \(X, Y, Z \leq 50\).

### Step 5: Change of Variables
To account for the upper limits, we can introduce new variables:
\[ X' = 50 - X, \quad Y' = 50 - Y, \quad Z' = 50 - Z \]
This transforms our equation into:
\[ (50 - X') + (50 - Y') + (50 - Z') = 101 \]
which simplifies to:
\[ 150 - (X' + Y' + Z') = 101 \quad \Rightarrow \quad X' + Y' + Z' = 49 \]

### Step 6: Count the Non-Negative Solutions
The number of non-negative integer solutions to the equation \(X' + Y' + Z' = 49\) can be found using the "stars and bars" theorem:
\[ \text{Number of solutions} = \binom{49 + 3 - 1}{3 - 1} = \binom{51}{2} \]

### Step 7: Calculate the Binomial Coefficient
Now, we compute:
\[ \binom{51}{2} = \frac{51 \times 50}{2} = 1275 \]

### Conclusion
Thus, the total number of ways to distribute the 101 coins among the three sons, ensuring that no one receives more than the combined total of the other two, is:
\[ \boxed{1275} \]