Question: How many ways can one distribute 

A) 15 Balls into 3 bags. Both bag and balls are distinct (labelled) and each bag must contain at least one ball.

B) 10 balls into 3 bags. again both bag and balls are distinct.

Thank you for your wisdom!

Regards

Answer: 
**Hint**

Try the following procedure:

1. Select how many balls for each bag, say $k\_1 + k\_2 + k\_3 = n$ where $n$ is your number of balls. How many ways are there to do that?
2. Now how many ways are there to select $k\_1$ balls from $n$ balls to go in the first bag?
3. How many ways to pick the $k\_2$ balls to go into the second bag from $n - k\_1$ balls?
4. How many ways to place remaining balls into the third bag?

Now, how many ways total could you perform this procedure in?
**First problem:** We use Inclusion/Exclusion. Call the bags A, B, and C.

There are $3^{15}$ ways to place the balls into bags, with no restriction. This is because for every one of the balls, we have $3$ choices for the bag it goes into.

However, every bag must contain at least one ball. So we count the number of **bad** ways to distribute the balls, that is, ways in which one or more bags remain empty.

There are $2^{15}$ distributions in which A is empty. Similarly, there are $2^{15}$ distributions in which B is empty, and $2^{15}$ in which C is empty.

If we calculate the sum $2^{15}+2^{15}+2^{15}$, we will have double-counted the $3$ distributions in which all the balls end up in the same bag.

For example, when we add together the $2^{15}$ for which A is empty, and the $2^{15}$ for which B is empty, we have double counted the distribution in which A and B are both empty, that is, all the balls end up in C.

So the number of bad distributions is $3\cdot 2^{15}-3$, and therefore the required number is $3^{15}-3\cdot 2^{15}+3$. 

**Second problem:** This one is much easier, since there appear to be no restrictions.