The number of ordered triplets, positive integers which are solutions of the equation x+y+z=100 is:

Options

A. 5081
B. 6005
C. 4851
D. 498

Answer

To find the number of ordered triplets of positive integers (x, y, z) that satisfy the equation \( x + y + z = 100 \), we can use the combinatorial method known as "stars and bars".

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the number of ordered triplets (x, y, z) such that their sum equals 100. Since x, y, and z are positive integers, each of them must be at least 1.

2. **Transforming the Variables**:
To simplify the problem, we can make a change of variables. Let:
\[ x' = x - 1, \quad y' = y - 1, \quad z' = z - 1 \]
Here, \( x', y', z' \) are non-negative integers (i.e., \( x', y', z' \geq 0 \)). The equation now becomes:
\[ (x' + 1) + (y' + 1) + (z' + 1) = 100 \]
Simplifying this gives:
\[ x' + y' + z' = 97 \]

3. **Applying the Stars and Bars Theorem**:
The problem now is to find the number of non-negative integer solutions to the equation \( x' + y' + z' = 97 \). According to the stars and bars theorem, the number of solutions is given by:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the total sum (97 in this case) and \( r \) is the number of variables (3 here).

4. **Substituting the Values**:
Here, \( n = 97 \) and \( r = 3 \). Therefore, we need to calculate:
\[ \binom{97 + 3 - 1}{3 - 1} = \binom{99}{2} \]

5. **Calculating \( \binom{99}{2} \)**:
The formula for combinations is:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus:
\[ \binom{99}{2} = \frac{99!}{2!(99-2)!} = \frac{99 \times 98}{2 \times 1} = \frac{9702}{2} = 4851 \]

6. **Conclusion**:
Therefore, the number of ordered triplets of positive integers (x, y, z) that satisfy the equation \( x + y + z = 100 \) is **4851**.