If x, y, z are integers such that `x >=0, y >=1, z >=2` and x + y + z = 15 , then the number of values of ordered triplets (x,y,z) are

Options

A. 91
B. 455
C. `""^(17)C_(15)`
D. 55

Answer

To solve the problem, we need to find the number of ordered triplets \((x, y, z)\) such that:

- \(x \geq 0\)
- \(y \geq 1\)
- \(z \geq 2\)
- \(x + y + z = 15\)

### Step 1: Transform the variables
Since \(y\) must be at least 1 and \(z\) must be at least 2, we can redefine the variables to simplify the problem:
- Let \(y' = y - 1\) (so \(y' \geq 0\))
- Let \(z' = z - 2\) (so \(z' \geq 0\))

Now we can express \(y\) and \(z\) in terms of \(y'\) and \(z'\):
- \(y = y' + 1\)
- \(z = z' + 2\)

### Step 2: Substitute into the equation
Substituting these new variables into the original equation \(x + y + z = 15\):
\[ x + (y' + 1) + (z' + 2) = 15 \]
This simplifies to:
\[ x + y' + z' + 3 = 15 \]
Subtracting 3 from both sides gives:
\[ x + y' + z' = 12 \]

### Step 3: Count the non-negative integer solutions
Now we need to find the number of non-negative integer solutions to the equation \(x + y' + z' = 12\). This is a classic problem in combinatorics that can be solved using the "stars and bars" theorem.

According to the stars and bars theorem, the number of ways to distribute \(n\) indistinguishable objects (stars) into \(r\) distinguishable boxes (variables) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case:
- \(n = 12\) (the total we want)
- \(r = 3\) (the number of variables: \(x\), \(y'\), and \(z'\))

### Step 4: Apply the formula
Using the formula:
\[ \binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} \]

### Step 5: Calculate \(\binom{14}{2}\)
Calculating \(\binom{14}{2}\):
\[ \binom{14}{2} = \frac{14 \times 13}{2 \times 1} = \frac{182}{2} = 91 \]

### Conclusion
Thus, the number of ordered triplets \((x, y, z)\) that satisfy the given conditions is **91**.

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