The number of non negative integral solution of the equation, `x+ y+3z =33` is

Options

A. 120
B. 135
C. 210
D. 520

Explanation

To find the number of non-negative integral solutions of the equation \(x + y + 3z = 33\), we can follow these steps:

### Step 1: Rewrite the equation
We can express \(x + y\) in terms of \(z\):
\[ x + y = 33 - 3z \]
This means that for each non-negative integer value of \(z\), \(x + y\) will take on a corresponding value.

### Step 2: Determine the range of \(z\)
Since \(x\) and \(y\) are non-negative integers, \(33 - 3z\) must also be non-negative. Therefore, we need:
\[ 33 - 3z \geq 0 \implies 3z \leq 33 \implies z \leq 11 \]
Thus, \(z\) can take values from \(0\) to \(11\) (inclusive).

### Step 3: Calculate the number of solutions for each value of \(z\)
For each fixed value of \(z\), the equation \(x + y = 33 - 3z\) can be solved using the stars and bars method. The number of non-negative integral solutions of the equation \(x + y = n\) is given by:
\[ \binom{n + r - 1}{r - 1} \]
where \(r\) is the number of variables (in this case, \(2\) for \(x\) and \(y\)). Thus, we have:
\[ \text{Number of solutions} = \binom{(33 - 3z) + 2 - 1}{2 - 1} = \binom{34 - 3z}{1} = 34 - 3z \]

### Step 4: Sum the solutions for all values of \(z\)
Now, we will sum the number of solutions for each integer value of \(z\) from \(0\) to \(11\):
\[ \text{Total solutions} = \sum_{z=0}^{11} (34 - 3z) \]
This can be simplified:
- When \(z = 0\), the number of solutions is \(34\).
- When \(z = 1\), the number of solutions is \(31\).
- When \(z = 2\), the number of solutions is \(28\).
- Continuing this way until \(z = 11\), where the number of solutions is \(1\).

### Step 5: Recognize the sequence
The sequence of solutions forms an arithmetic progression (AP) where:
- First term \(a = 34\)
- Last term \(l = 1\)
- Number of terms \(n = 12\) (from \(z = 0\) to \(z = 11\))

### Step 6: Calculate the sum of the arithmetic progression
The sum \(S_n\) of the first \(n\) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (a + l) \]
Substituting the values:
\[ S_{12} = \frac{12}{2} \times (34 + 1) = 6 \times 35 = 210 \]

### Final Answer
Thus, the total number of non-negative integral solutions of the equation \(x + y + 3z = 33\) is:
\[ \boxed{210} \]