Distributing 6 oranges, 1 apple, 1 banana and 1 pineapple among 3 children if we were to have 6 oranges, 1 apple, 1 banana and 1 pineapple. On how many ways can we distribute this to 3 children (each child must receive at least one fruit)? 
I was trying to do it in a way where we say that we have 9 oranges which we can distribute 28 ways and that we multiply this with 3 for every different types of fruit, except oranges like: 28*3*3*3 but this isn't correct. 
 A: This problem is complicated a bit by the condition that every child gets at least 1 fruit. Without that condition, the solution is computed as follows:
There are $\binom{6+3-1}{3-1} = \binom{8}{2} = 28$ ways to distribute the 6 oranges. For all the other fruits, there are $3$ possibilities. So we get the number
$$28\cdot 3\cdot 3\cdot 3 = 756.$$
Now lets modify this solution for the original problem. The above number is not correct, since it includes the forbidden distributions where some child doesn't get any fruit.
To compensate for that, we see that there are $\binom{7}{1}\cdot 2\cdot 2\cdot 2 = 56$
distributions where a fixed child doesn't get any fruit. This leads to the new number
$$756 - 3\cdot 56 = 588.$$
This still isn't correct, since we subtracted those cases twice where two fixed childs don't get any fruits. For each pair of childs there is a single such distribution (the third kid gets all the fruits).
So the final answer is
$$ 588 + 3 = \mathbf{591}.$$
The above reasoning is called "principle of inclusion and exclusion".
A: azimut has provided an elegant solution.  Here is another method:
Consider the ways in which we can distribute the apple, banana, and pineapple.  


*

*We can give each child one of these three fruits, which can be done in $3! = 6$ ways. Since each child now has a fruit, the number of ways we can distribute the oranges is the number of solutions of the equation 
$$x_1 + x_2 + x_3 = 6$$
in the non-negative integers, which is equivalent to the number of ways we can place two addition signs in a list of six ones, which is $$\binom{6 + 2}{2} = \binom{8}{2} = 28$$  Hence, the number of ways we can distribute the fruit if each child receives one fruit selected from the apple, banana, and pineapple is 
$$3!\binom{8}{2} = 6 \cdot 28 = 168$$ 

*We can give one child two fruits selected from the apple, banana, and pineapple, and the third of these three fruits to one of the two remaining children, which can be done in $$3 \cdot \binom{3}{2} \cdot 2 = 18$$ ways since there are three ways to select the child who receives three pieces of fruit, $\binom{3}{2}$ ways of selecting the fruits that child receives, and two ways of selecting the child who receives the remaining piece of fruit. When we distribute the oranges, the third child must receive an orange, so we are left with five oranges to distribute among the three children, which can be done in $$\binom{5 + 2}{2} = \binom{7}{2} = 21$$ ways.  Hence, there are 
$$3 \cdot \binom{3}{2} \cdot 2 \cdot \binom{7}{2} = 18 \cdot 21 = 378$$
ways to distribute the fruit when we give two fruits selected from the apple, banana, and pineapple to one child and the third fruit to a different child.  

*We can give one of the three children the apple, banana, and pineapple, which can be done in three ways since there are three children.  We must give each of the other two children an orange, leaving us with four oranges to distribute, which can be done in $$\binom{4 + 2}{2} = \binom{6}{2} = 15$$ ways.  Thus, the number of ways we can distribute the fruit if we give one child the apple, banana, and pineapple is 
$$3 \cdot \binom{6}{2} = 3 \cdot 15 = 45$$ 


Adding the three disjoint cases yields a total of $168 + 378 + 45 = 591$ ways to distribute the fruit among the children if each child receives at least one fruit.
