In how many ways can we distribute 20 SAR among 3 children, the two youngest children get at least 5 SAR but no more than 10 SAR

We are asked to distribute 20 SAR among three children such that the two youngest children receive at least 5 SAR but no more than 10 SAR each. Let's define the children as \( C_1, C_2, C_3 \), with \( C_2 \) and \( C_3 \) being the two youngest.

### Step-by-Step Breakdown:

1. **Constraints for \( C_2 \) and \( C_3 \):**
   - Both \( C_2 \) and \( C_3 \) must receive between 5 and 10 SAR.
   - Let \( x_2 \) and \( x_3 \) represent the amount of SAR that \( C_2 \) and \( C_3 \) receive, respectively.
   - Therefore, the constraints are:
     \[
     5 \leq x_2 \leq 10 \quad \text{and} \quad 5 \leq x_3 \leq 10.
     \]
   
   To simplify this, define new variables \( y_2 = x_2 - 5 \) and \( y_3 = x_3 - 5 \). This transforms the constraints to:
   \[
   0 \leq y_2 \leq 5 \quad \text{and} \quad 0 \leq y_3 \leq 5.
   \]
   Now, \( y_2 \) and \( y_3 \) represent the "extra" SAR beyond 5 SAR that the two youngest children receive.

2. **Total Amount for \( C_1 \):**
   - The total SAR to be distributed is 20.
   - Therefore, the amount that \( C_1 \) (the oldest child) receives is:
     \[
     x_1 = 20 - (x_2 + x_3) = 20 - ((y_2 + 5) + (y_3 + 5)) = 10 - (y_2 + y_3).
     \]
   - So the amount \( C_1 \) receives, \( x_1 = 10 - (y_2 + y_3) \), must be non-negative. Hence, we require:
     \[
     y_2 + y_3 \leq 10.
     \]

3. **Number of Solutions:**
   - We now need to count the number of non-negative integer solutions to the equation \( y_2 + y_3 \leq 5 \), where \( 0 \leq y_2 \leq 5 \) and \( 0 \leq y_3 \leq 5 \).
   - This is a classic case of counting the number of integer solutions to an inequality, which can be done using the "stars and bars" method.
   - The number of solutions to \( y_2 + y_3 = k \), where \( 0 \leq k \leq 5 \), is \( k + 1 \) (since \( y_2 \) can range from 0 to \( k \), and \( y_3 \) is determined by the value of \( k \)).
   
   So, the total number of solutions is:
   \[
   \sum_{k=0}^{5} (k + 1) = 1 + 2 + 3 + 4 + 5 + 6 = 21.
   \]

### Final Answer:
There are **21 ways** to distribute 20 SAR among the three children, with the two youngest receiving between 5 and 10 SAR each.

Would you like more details on how this method works or any clarifications?

Here are 5 related questions to explore:

1. How does the "stars and bars" method work in combinatorics?
2. What if there were 4 children with similar constraints?
3. How would the solution change if \( C_2 \) and \( C_3 \) could receive between 5 and 15 SAR?
4. What if we change the total amount of SAR to 25? How many ways then?
5. What if \( C_1 \) also had constraints on the amount they can receive?

**Tip:** When solving combinatorics problems, always try simplifying constraints using new variables to make calculations easier.

In how many ways can we distribute 20 SAR among 3 children, the two youngest children get at least 5 SAR but no more than 10 SAR?

Discover how to distribute 20 SAR among three children, ensuring the two youngest children receive between 5 and 10 SAR each. Learn about using the stars and bars method to find that there are 21 possible ways to distribute the money. This combinatorics problem is ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation