Estonian Mathematical Olympiad

According to the conditions, each child must receive 9 fruits. It suffices to find how many possibilities there are to give the first child 9 fruits, because the second child receives all the remaining fruits. The first child can be given 0 to 6 pears and 0 to 5 oranges. Disregarding the condition that he must receive 9 fruit in total, there are a total of $7\cdot 6$, or 42, possibilities for giving pears and oranges. The possibilities where the first child receives more than 9 of pears and oranges alone, and also those where the first child gets less than 2 of pears and oranges are not suitable. The possibilities where the total number of pears and oranges is more than 9 are 3 (6 + 4, 5 + 5 and 6 + 5), while the possibilities where the total number of pears and oranges is less than 2 are also 3 (1 + 0, 0 + 1 and 0 + 0). Thus, there are $42-3-3=36$ suitable possibilities.

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Alternative solution.

Again it suffices to find how many possibilities there are to give the first child 9 fruits. If the first child gets more apples than the second child, then the first child gets 4 apples and 5 more fruits. Mark with 5 circles the fruits – apples, followed by pears, and finally oranges – and with 2 dashes the places where one kind of fruit changes to another. Then all possible choices of 5 fruits are represented as a word consisting of 7 characters, and the choice consists in determining the positions where the dashes are located. We get $\frac{7\cdot 6}{2}=21$ possibilities. But the possibilities where before the first dash there are more than 3 circles are not suitable, because we have only 3 apples left. In this case the dashes are either on fifth and sixth, fifth and seventh or sixth and seventh position. Hence $21-3=18$ possibilities remain. The possibilities where the first child gets fewer apples than the second child are symmetrical, so there are the same number of them. So there are a total of $18\cdot 2=36$ different ways to distribute the fruits.