$N$ boxes are lined up in a sequence. You have $A$ red balls and $B$ blue balls. The red balls (and the blue ones) are exactly the same. You can place the balls in the boxes. It is allowed to put in a box, balls of the two kinds, or only from one kind. You can also leave some of the boxes empty. It's not necessary to place all the balls in the boxes. Count the number of different ways to place the balls in the boxes in the described way.
If there was only $1$ ball the answer would have been $\binom {N+A-1} {N-1}$. But I dont know how to solve it for $2$ balls.