How many different integer value solutions are there to the
following equations:
(a) (2 points) x1 + x2 + x3 + x4 + x5 = 25, xi > 0
(b) (2 points) x1 + x2 + x3 = 15, xi ≥ 0

Let's address both parts step by step:

### Part (a): \( x_1 + x_2 + x_3 + x_4 + x_5 = 25 \), where \( x_i > 0 \)

To solve this, we use the "stars and bars" method, which is commonly used to count the number of ways to distribute indistinguishable objects (stars) into distinct groups (bars).

Since \( x_i > 0 \), we can first transform the variables by setting \( y_i = x_i - 1 \), where \( y_i \geq 0 \). This way, the equation becomes:

\[
y_1 + y_2 + y_3 + y_4 + y_5 = 25 - 5 = 20
\]

Now, we need to count how many non-negative integer solutions exist for this equation. The number of non-negative integer solutions to an equation of the form:

\[
y_1 + y_2 + \cdots + y_k = n
\]

is given by the binomial coefficient \( \binom{n+k-1}{k-1} \). In our case, \( n = 20 \) and \( k = 5 \), so the number of solutions is:

\[
\binom{20+5-1}{5-1} = \binom{24}{4}
\]

Now, let's calculate \( \binom{24}{4} \):

\[
\binom{24}{4} = \frac{24 \times 23 \times 22 \times 21}{4 \times 3 \times 2 \times 1} = 10626
\]

Thus, there are **10,626** different integer value solutions for part (a).

---

### Part (b): \( x_1 + x_2 + x_3 = 15 \), where \( x_i \geq 0 \)

Here, we again use the "stars and bars" method, but this time with non-negative integers \( x_i \geq 0 \).

The number of non-negative integer solutions to this equation is given by the binomial coefficient \( \binom{n+k-1}{k-1} \), where \( n = 15 \) and \( k = 3 \).

\[
\binom{15+3-1}{3-1} = \binom{17}{2}
\]

Now, let's calculate \( \binom{17}{2} \):

\[
\binom{17}{2} = \frac{17 \times 16}{2 \times 1} = 136
\]

Thus, there are **136** different integer value solutions for part (b).

---

### Summary:
- (a) There are **10,626** solutions where \( x_1 + x_2 + x_3 + x_4 + x_5 = 25 \) with \( x_i > 0 \).
- (b) There are **136** solutions where \( x_1 + x_2 + x_3 = 15 \) with \( x_i \geq 0 \).

Would you like more details or clarifications?

Here are 5 related questions:
1. How does the stars and bars method work in combinatorics?
2. Can the stars and bars method handle equations with more complex constraints, such as \( x_i \leq n \)?
3. How would the solution change if there were upper limits on \( x_i \)?
4. What happens if \( x_i \) are required to be even integers?
5. How would we solve the equations if \( x_i \) must be prime numbers?

**Tip**: The stars and bars method is powerful for partitioning problems. Always check if the constraints on variables are strict (e.g., \( x_i > 0 \)) or flexible (e.g., \( x_i \geq 0 \)).

How many different integer value solutions are there to the following equations: (a) x1 + x2 + x3 + x4 + x5 = 25, xi > 0 (b) x1 + x2 + x3 = 15, xi ≥ 0

Explore how to find the number of integer solutions to equations like x1 + x2 + x3 + x4 + x5 = 25 and x1 + x2 + x3 = 15 using the Stars and Bars combinatorics method. This solution is suitable for high school students learning combinatorics and algebra, with detailed calculations and explanations.

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