If x, y and z are whole numbers such that X>= Y , then how many solutions are possible for the equation x + y + z = 36?

Options

A. 361
B. 323
C. 382
D. 342

Answer

To solve the problem of finding the number of solutions for the equation \(x + y + z = 36\) under the condition that \(x \geq y\), we can follow these steps:

### Step 1: Total Solutions Without Constraints
First, we will find the total number of solutions to the equation \(x + y + z = 36\) without any constraints.

Using the stars and bars method, we need to place 36 stars (representing the total sum) and 2 bars (to separate \(x\), \(y\), and \(z\)). The total number of symbols (stars + bars) is \(36 + 2 = 38\).

The number of ways to arrange these symbols is given by the combination formula:

\[ \text{Total combinations} = \binom{38}{2} \]

Calculating this:

\[ \binom{38}{2} = \frac{38 \times 37}{2} = 703 \]

### Step 2: Configuration Where \(x = y\)
Next, we consider the case where \(x = y\). In this case, we can rewrite the equation as:

\[ 2x + z = 36 \]

From this, we can express \(z\) in terms of \(x\):

\[ z = 36 - 2x \]

To ensure \(z\) remains a whole number, \(36 - 2x \geq 0\) must hold, which leads to:

\[ x \leq 18 \]

Thus, \(x\) can take values from \(0\) to \(18\), giving us \(19\) possible values for \(x\) (including \(0\)).

### Step 3: Configuration Where \(x < y\) and \(x > y\)
Now, we need to find the configurations where \(x < y\) and \(x > y\). The total configurations without any constraints is \(703\), and we subtract the configurations where \(x = y\):

\[ \text{Configurations where } x < y \text{ or } x > y = 703 - 19 = 684 \]

Since the cases \(x < y\) and \(x > y\) are symmetric, we can divide this result by \(2\):

\[ \text{Configurations where } x > y = \frac{684}{2} = 342 \]

### Step 4: Total Configurations Where \(x \geq y\)
Finally, to find the total configurations where \(x \geq y\), we add the configurations where \(x = y\) to those where \(x > y\):

\[ \text{Total configurations where } x \geq y = 342 + 19 = 361 \]

### Conclusion
Thus, the total number of solutions for the equation \(x + y + z = 36\) under the condition that \(x \geq y\) is:

\[ \boxed{361} \]