Total number of integral solutions of the system of equations `x_(1)+x_(2)+x_(3)+x_(4)+x_(5)=20and x_(1)+x_(2)+x_(3)=5" when "x_(k)gel0,"is"`

Options

A. 334
B. 336
C. 332
D. 338

Answer

To find the total number of integral solutions for the given system of equations:

1. **Write down the equations:**
\[ x_1 + x_2 + x_3 + x_4 + x_5 = 20 \quad \text{(1)} \]
\[ x_1 + x_2 + x_3 = 5 \quad \text{(2)} \]

2. **Subtract equation (2) from equation (1):**
\[ (x_1 + x_2 + x_3 + x_4 + x_5) - (x_1 + x_2 + x_3) = 20 - 5 \]
This simplifies to:
\[ x_4 + x_5 = 15 \quad \text{(3)} \]

3. **Now we need to find the number of non-negative integral solutions to equation (3):**
The equation \(x_4 + x_5 = 15\) can be solved using the stars and bars method. The formula for the number of non-negative integral solutions of the equation \(x_1 + x_2 + ... + x_k = n\) is given by:
\[ \binom{n + k - 1}{k - 1} \]
where \(n\) is the total and \(k\) is the number of variables.

Here, \(n = 15\) and \(k = 2\) (for \(x_4\) and \(x_5\)). Thus, we have:
\[ \text{Number of solutions} = \binom{15 + 2 - 1}{2 - 1} = \binom{16}{1} \]

4. **Calculate \(\binom{16}{1}\):**
\[ \binom{16}{1} = 16 \]

5. **Now, we also need to find the number of non-negative integral solutions for equation (2) \(x_1 + x_2 + x_3 = 5\):**
Using the same formula:
\[ n = 5, \quad k = 3 \]
\[ \text{Number of solutions} = \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} \]

6. **Calculate \(\binom{7}{2}\):**
\[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]

7. **Finally, multiply the number of solutions from both equations:**
\[ \text{Total solutions} = 16 \times 21 = 336 \]

Thus, the total number of integral solutions of the system of equations is **336**.

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