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.