Distributing black and white marbles...

There are six identical white marbles and eight identical black marbles. In how many ways can we give these marbles to 3 boys, if every boy must get at least one marble, and we don't necessarily have to give them all of the marbles?

This problem should be solved using the inclusion-exclusion principle.

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I tried to solve this by defining three sets $\displaystyle A_1, A_2, A_3$ where each set $\displaystyle A_i$ contains all the possible marble combinations (or sets of marbles) $\displaystyle i$-th boy can get, and then counting the number of ordered triples $\displaystyle (x,y,z)$ where $\displaystyle x \in A_1, y \in A_2, z \in A_3$. However - this approach is wrong, because with the sets thus defined it's possible to have a triplet $\displaystyle (x,y,z)$ representing the first boy getting e.g. 8 black marbles, second boy getting 7 black marbles and third boy 6 black marbles, which is not possible as there are only 8 black marbles...

So I'd be immensely grateful for any help/hints/insights you might have :D