Find the number of non negative integral solutions of:
` x_1 +x_2 +x_3 +4x_4=20`

Options

A. a. 536
B. b. `""^(43)C_3`
C. c. 666
D. d. none of these

Answer

To find the number of non-negative integral solutions to the equation \( x_1 + x_2 + x_3 + 4x_4 = 20 \), we can use the method of generating functions or the stars and bars theorem. Here’s a step-by-step solution:

### Step 1: Rewrite the equation
We start with the equation:
\[ x_1 + x_2 + x_3 + 4x_4 = 20 \]
This equation can be interpreted as distributing 20 units among the variables \( x_1, x_2, x_3 \), and \( x_4 \) where \( x_4 \) contributes four times its value to the total.

### Step 2: Change the variable for \( x_4 \)
To simplify the equation, we can introduce a new variable \( y = 4x_4 \). This means we can express \( x_4 \) in terms of \( y \):
\[ x_4 = \frac{y}{4} \]
Now, substituting \( y \) into the equation gives:
\[ x_1 + x_2 + x_3 + y = 20 \]
where \( y \) must be a multiple of 4 (i.e., \( y = 0, 4, 8, 12, 16, 20 \)).

### Step 3: Count the number of solutions for each case of \( y \)
We will consider the different possible values of \( y \):

1. **If \( y = 0 \)**:
\[ x_1 + x_2 + x_3 = 20 \]
The number of non-negative integer solutions is given by the stars and bars theorem:
\[ \binom{20 + 3 - 1}{3 - 1} = \binom{22}{2} = 231 \]

2. **If \( y = 4 \)**:
\[ x_1 + x_2 + x_3 = 16 \]
The number of solutions is:
\[ \binom{16 + 3 - 1}{3 - 1} = \binom{18}{2} = 153 \]

3. **If \( y = 8 \)**:
\[ x_1 + x_2 + x_3 = 12 \]
The number of solutions is:
\[ \binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} = 91 \]

4. **If \( y = 12 \)**:
\[ x_1 + x_2 + x_3 = 8 \]
The number of solutions is:
\[ \binom{8 + 3 - 1}{3 - 1} = \binom{10}{2} = 45 \]

5. **If \( y = 16 \)**:
\[ x_1 + x_2 + x_3 = 4 \]
The number of solutions is:
\[ \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15 \]

6. **If \( y = 20 \)**:
\[ x_1 + x_2 + x_3 = 0 \]
The number of solutions is:
\[ \binom{0 + 3 - 1}{3 - 1} = \binom{2}{2} = 1 \]

### Step 4: Sum all the solutions
Now we sum all the solutions for each case of \( y \):
\[ 231 + 153 + 91 + 45 + 15 + 1 = 536 \]

### Final Answer
Thus, the total number of non-negative integral solutions to the equation \( x_1 + x_2 + x_3 + 4x_4 = 20 \) is **536**.

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