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3. Use the generating functions to model and find the solutions for the following problems:   b. In how many ways can we distribute 20 SAR among 3 children, if the oldest child should have at least 10 SAR?

To solve the problem of distributing 20 SAR among three children, where the oldest child must receive at least 10 SAR, we can use generating functions. The image you've provided contains steps related to this type of problem. Let's break it down:

### Problem:
- You are asked to distribute 20 SAR among three children such that the oldest child gets at least 10 SAR.
  
### Solution Steps:
1. **Oldest Child's Share**: 
   - Since the oldest child must have at least 10 SAR, we can remove these 10 SAR from the total, effectively reducing the problem to distributing the remaining **\(20 - 10 = 10\) SAR** among the three children, but now the oldest child can receive any non-negative amount of SAR.

2. **Generating Function**:
   - The generating function for distributing money without any constraints is typically of the form:
     \[
     \frac{1}{1 - x} 
     \]
     where \(x\) represents the number of SAR distributed. 

3. **Adjustments for the Oldest Child**:
   - Since the oldest child has already received 10 SAR, we factor that out. The distribution of the remaining amount is modeled by the generating function:
     \[
     x^{10} \cdot \left(\frac{1}{1 - x}\right)^3
     \]
     Here, \(x^{10}\) accounts for the 10 SAR already given to the oldest child, and the term \(\left(\frac{1}{1 - x}\right)^3\) models the distribution of the remaining 10 SAR among all three children.

4. **Expansion of the Generating Function**:
   - To solve for the coefficient of \(x^{20}\) (which gives the number of ways to distribute 20 SAR), we expand:
     \[
     x^{10} \cdot \left(\frac{1}{1 - x}\right)^3
     \]
     Using the binomial series, the expansion of \(\frac{1}{(1 - x)^3}\) is given by:
     \[
     \sum_{k=0}^{\infty} \binom{k+2}{2} x^k
     \]
     The required term is the coefficient of \(x^{10}\) in this expansion, which would give us the solution.

5. **Final Answer**:
   - From the series expansion, the coefficient of \(x^{10}\) represents the number of ways to distribute the remaining 10 SAR among the children. This corresponds to the binomial coefficient calculation:
     \[
     \binom{10 + 2}{2} = \binom{12}{2} = 66
     \]
   - Therefore, there are 66 ways to distribute 20 SAR among three children, where the oldest child gets at least 10 SAR.

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### Do you need any clarifications or further steps?

Here are 5 related questions you might find helpful:
1. What are generating functions, and how do they model distribution problems?
2. How can generating functions be applied to problems with constraints?
3. What is the binomial series expansion, and how is it used in combinatorics?
4. How would the solution change if the oldest child needed a different minimum amount of SAR?
5. Can generating functions be used for problems involving other types of constraints?

**Tip:** When dealing with problems involving generating functions, always identify the constraint and adjust the generating function accordingly before expanding.

In how many ways can we distribute 20 SAR among 3 children, if the oldest child should have at least 10 SAR?

Explore how to distribute 20 SAR among 3 children using generating functions, where the oldest child must receive at least 10 SAR. Learn how to apply combinatorics, generating functions, and binomial expansions to solve this distribution problem.

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