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.