Count the number of ways to place the identical balls in distinct bins We have $6$ balls and $4$ bins.
a) 
Count the number of ways to place the identical balls in identical bins
b) 
Count the number of ways to place the identical balls in distinct bins
a) My  solution looks like this:


*

*$(6,0,0,0)$

*$(2,4,0,0)$

*$(3,3,0,0)$

*$(5,1,0,0)$

*$(2,3,1,0)$

*$(2,2,2,0)$

*$(4,1,1,0)$

*$(3,1,1,1)$

*$(2,1,1,2)$ 
And $9$ is the correct answer: 
b) However im unsure of to think here. I guess for the first case $(6,0,0,0)$ there’s now 4 different ways to place the balls if the bins are distinct? But how many ways would there be for $(2,4,0,0)$, $(2,1,1,2)$ and so on? Is there a way to think here so that we can solve this problem? 
$84$ should be the correct answer for b).
 A: I don't think it is easy to go from (a) to (b) or vice versa (unless the numbers are very small).
Part (b) is much easier.  It is solved by stars and bars and has a closed-form solution (in fact, a binomial coefficient).
For part (a), we're talking about integer partitions but with the additional restriction that the number of parts $\le 4$.  The wiki article includes the restricted case where the number of parts is exactly $k$, so I suppose you can sum over $k \in [1, 4]$.  Anyway, I don't think there is a general solution.  Here is a related MSE thread which limits both the number of parts to $\le k$ and the size of each part to $\le d$.  So if we set $d=c$ so that each part is unlimited but the number of parts is still limited to $\le k$ then this becomes the OP question.
A: HINT: For $(4,1,1,0)$ case, we can choose the empty bin with $\binom{4}{1}$ and since balls are identical, we can put them to three bins with $\frac{3!}{2!} = 3$. Here, we are dividing it by $2!$ because balls are identical and we are putting same number of balls into two bins. So, in this specific case, there are $\binom{4}{1}\cdot\frac{3!}{2!} = 12$ different ways.
Now, can you find how many different ways in other cases?
A: I have a solution for part (b):
Let the number of balls in each of the 4 distinct bins be $x_1$, $x_2$, $x_3$ and $x_4$ respectively. The total number of balls (let the balls be represented by the character $\phi$) is $6\phi$ (as there are 6 balls, and each ball is $\phi$), therefore we can form the following equation-
$$x_1+x_2+x_3+x_4=6\phi$$
Solving this is straightforward- in how many ways can we write $6$ as the sum of 4 non-negative integers?
This is one way to solve it-
Let each ball be represented by the character $\phi$. Consider one case where there are 3 balls in the first bin, 1 ball in the second, third and the fourth bins, each. When I put this in the above equation:
$$\phi\phi\phi +\phi+\phi+\phi=6\phi$$
Note that we are NOT multiplying anything here.
The question now has simplified to- how many ways can we permute the elements of the following set: 
$$\{\phi,\phi,\phi,\phi,\phi,\phi,+,+,+\}$$
This is one way of finding all the ways we can move the balls ($\phi$) into different boxes; i.e. we are just moving around the $\phi$'s and $+$'s in the above equation.
The answer is: 
$$\dfrac{9!}{6!\times3!}=84.$$
