SOLUTION: In how many ways can we distribute seven bananas and six oranges among four children so that each child receives at least one banana and at least one orange?

If  C1, C2, C3, C4 are the four children, we can draw their minimum allotment of bananas and oranges as follows:

               C1      C2      C3      C4
Bananas         1       1       1       1
Oranges         1       1       1       1

That leaves  3 bananas and 2 oranges to be distributed to the 4 children.   So the original problem is equivalent to "how many ways can 3 bananas and 2 oranges be distributed to 4 children,  where each child can receive 0…3 bananas and 0…2 oranges ?"

3 bananas:  
All 3 can go to one child in     4 ways.
If we give 2 to any one child (4 ways) then the remaining one can be distributed in 3 ways (4x3=12 ways total).
If we give 1 banana to each of 3 children, that can be done in 4 ways

Total number of ways to distribute the bananas is 4+12+4 = 20 ways.

2 oranges:
We can give both oranges to one child in 4 ways.
We can give one orange to one child and the other orange to another child in 6 ways (OO—, -OO-, —OO, O-O-, -O-O, and O—O,  where O=orange, -=nothing)

Total number of ways to distribute the oranges is 4+6 = 10 ways.