If each marble is different?
You must decide between trhee jars for each marble, so $3^{15}$
If each marble is the same?
You must decide how many to put in A, then how many in B (the rest go in C). So:
being $N = 15$
- how many in A: $A$ balls ($0\le A\le 15$), N+1 posibilities
- how many in B: $B$ balls ($0\le B\le 15-A$), N-A+1 posibilities
So, the answer will be
$\sum_{A=0}^{N}{(N-A+1)} = $ $ = (N+1)(N+1) - \sum_{A=0}^{N}{A} = $
$ = N(N+1) + (N+1) - \frac{N(N+1)}{2} = $
$ = \frac{N(N+1)}{2} + (N+1) = $
$ = \frac{N(N+1) + 2(N+1)}{2} = $
$ = \frac{(N+2)(N+1)}{2} $
If each marble is the same and each jar must have at least two marbles?
Put two marbles in each jar. Then you have 9 marbles left. So answer will be like (b) but with $N = 9$
If each marble is the same but each jar can have at most 6 marbles?
Put 6 marbles in each jar. Now you must take from them 3 balls to get a valid solution. You can pick how many balls to take from each jar. how many (0 to 3) from A, and then how many from B (0 to 3-A). That's analogous to the previous problem, but with three balls.
$\frac{(N+1)(N+2)}{2}$
where $N = 3$
If you have 10 identical red marbles and 5 identical blue marbles?
Place first red marbles, then blue marbles. They are independent so the number of posibilities will be the number posibilities for placing red marbles multiplied by the number of posibilities for placing the blue marbles.
$\frac{(Na+1)(Na+2)}{2} . \frac{(Nb+1)(Nb+2)}{2} $
$Na = 10, Nb = 5$