I don't understand how the stars and bars method is different from choosing k-1 spaces between n objects which have n-1 spaces, so the equation should be n-1 choose k-1?

The confusion you're facing comes from the distinction between the "stars and bars" method and selecting spaces between objects. Let's break it down carefully.

### Stars and Bars Method

The "stars and bars" method is a way to solve problems where you need to divide \( n \) identical objects (stars) into \( k \) distinct groups (bins). The equation for this is:
\[
\binom{n + k - 1}{k - 1}
\]
where:
- \( n \) is the number of stars (objects),
- \( k \) is the number of bins (groups),
- The \( k - 1 \) comes from the fact that we need \( k - 1 \) bars to divide the \( n \) objects into \( k \) groups.

### Choosing \( k-1 \) Spaces Between \( n \) Objects

Now, if you're thinking of the case where you are selecting \( k - 1 \) spaces between \( n \) objects, this would be a **different type of problem**. In this scenario:
- You have \( n \) objects (not identical), and you're choosing \( k-1 \) positions between them.
- There are \( n - 1 \) possible spaces (because there are \( n \) objects), so the correct equation here would indeed be \( \binom{n - 1}{k - 1} \).

### Difference Between the Two Methods

1. **Stars and Bars**: You are dealing with dividing identical objects (stars) into \( k \) distinct groups. The total number of available positions for the \( k-1 \) dividers is increased by the number of objects because you're considering both stars and dividers. That's why the equation is \( \binom{n + k - 1}{k - 1} \).

2. **Choosing Spaces Between Objects**: Here, you're selecting actual gaps between distinct objects. You're not trying to divide identical objects into groups. Instead, you have a fixed number of gaps, and you're choosing a certain number of them. That's why you use \( \binom{n - 1}{k - 1} \) in this case.

### Example to Clarify
#### Stars and Bars Example:
Suppose you want to divide 5 identical stars into 3 groups. You could think of the stars as "S" and the bars as "|". For instance, one possible arrangement is:
\[
S \, S \, | \, S \, | \, S \, S
\]
This divides the stars into 3 groups: {2, 1, 2}. Here, you have 5 stars and need 2 bars, so the total number of objects to arrange is \( 5 + 2 = 7 \). Hence, you calculate \( \binom{7}{2} \).

#### Choosing Spaces Example:
Suppose you have 5 distinct objects in a line, and you want to choose 2 spaces to divide them into 3 groups. The possible gaps between the objects are fewer than the total objects (since the objects themselves aren't identical). There are 4 spaces, and you choose 2, so the calculation would be \( \binom{4}{2} \).

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Let me know if you want to dive deeper into a specific example! Here are five related questions to explore further:

1. How would you apply the stars and bars method to divide 7 identical objects into 4 groups?
2. What happens if the objects you're dividing are distinct rather than identical?
3. Can you apply the stars and bars method when some of the groups are allowed to be empty?
4. How does the formula \( \binom{n + k - 1}{k - 1} \) change if we restrict the groups to be non-empty?
5. How does this concept relate to integer partitions?

**Tip**: When using the stars and bars method, the number of groups (or bins) is what determines how many dividers (bars) you need, not the number of objects!

Explore the distinction between the stars and bars method for dividing n identical objects into k groups and choosing spaces between distinct objects. This article provides a detailed explanation, including the relevant binomial coefficients and combinatorial techniques. Suitable for Grades 10-12.

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