Number of integral solutions of `2x+y+z=10 (xge0,yge0,Zge0)` is

Options

A. 18
B. 27
C. 36
D. 51

Answer

To find the number of integral solutions for the equation \(2x + y + z = 10\) under the constraints \(x \geq 0\), \(y \geq 0\), and \(z \geq 0\), we can consider two cases based on the parity (even or odd) of the variables \(y\) and \(z\).

### Step 1: Case 1 - Both \(y\) and \(z\) are even
1. Let \(y = 2y'\) and \(z = 2z'\) where \(y'\) and \(z'\) are non-negative integers.
2. Substitute these into the equation:
\[ 2x + 2y' + 2z' = 10 \]
Simplifying gives:
\[ x + y' + z' = 5 \]
3. Now, we need to find the number of non-negative integral solutions to the equation \(x + y' + z' = 5\).
4. The formula for the number of non-negative integral solutions of the equation \(x_1 + x_2 + \ldots + x_r = n\) is given by:
\[ \binom{n + r - 1}{r - 1} \]
Here, \(n = 5\) and \(r = 3\) (for \(x\), \(y'\), and \(z'\)):
\[ \text{Number of solutions} = \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} = 21 \]

### Step 2: Case 2 - Both \(y\) and \(z\) are odd
1. Let \(y = 2y' + 1\) and \(z = 2z' + 1\) where \(y'\) and \(z'\) are non-negative integers.
2. Substitute these into the equation:
\[ 2x + (2y' + 1) + (2z' + 1) = 10 \]
Simplifying gives:
\[ 2x + 2y' + 2z' + 2 = 10 \implies 2x + 2y' + 2z' = 8 \implies x + y' + z' = 4 \]
3. Now, we need to find the number of non-negative integral solutions to the equation \(x + y' + z' = 4\).
4. Using the same formula as before:
\[ \text{Number of solutions} = \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15 \]

### Step 3: Total Solutions
1. Add the solutions from both cases:
\[ \text{Total solutions} = 21 + 15 = 36 \]

Thus, the total number of integral solutions for the equation \(2x + y + z = 10\) with the given constraints is **36**.

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