Answer:
there are 120 different ways to distribute 20 apples and 10 oranges to 3 children, with each child having at least 2 apples and at least one orange.
Step-by-step explanation:
To find the number of ways to distribute 20 apples and 10 oranges to 3 children, with each child having at least 2 apples and at least one orange, we can use the concept of stars and bars.
Let's denote the number of apples each child receives as A1, A2, and A3, and the number of oranges each child receives as O1, O2, and O3.
We can start by giving each child the minimum required apples and oranges: 2 apples and 1 orange.
So, we are left with 14 apples and 7 oranges to distribute among the three children.
To distribute the remaining apples, we can use the stars and bars technique. We have 14 apples and 2 dividers (to separate the apples between the children). This can be visualized as placing 14 stars and 2 bars in a row:
\* \* \* \* \* \* \* \* \* \* \* \* \* | |
Each configuration of stars and bars represents a different distribution of the remaining apples. The number of ways to arrange the stars and bars is given by the binomial coefficient:
C(n+k-1, k-1)
In this case, n = 14 (number of apples) and k = 2 (number of dividers/bars). So, we have:
C(14+2-1, 2-1) = C(15, 1) = 15
This means there are 15 different ways to distribute the remaining apples.
Now, we can consider the oranges. Each child must receive at least one orange, so we need to distribute the remaining 7 oranges among the three children.
Using the same stars and bars technique, we have 7 oranges and 2 dividers. This gives us:
C(7+2-1, 2-1) = C(8, 1) = 8
There are 8 different ways to distribute the remaining oranges.
To find the total number of ways to distribute both the apples and oranges, we multiply the number of ways for each:
15 (ways to distribute apples) * 8 (ways to distribute oranges) = 120
Therefore, there are 120 different ways to distribute 20 apples and 10 oranges to 3 children, with each child having at least 2 apples and at least one orange.
