A total of $60$ (indistinguishable) iPads are to be distributed to $7$ elementary school classrooms. How many ways are there to do this if 
A total of $60$ (indistinguishable) iPads are to be distributed to $7$ elementary school classrooms. How many ways are there to do this if


(a) There are no restrictions?


(b) Four of the classrooms receive iPads but three do not.

If the iPads were distinguishable then I could say that for (a) the answer would be $ = 60 \cdot 59 \cdot ... \cdot 54$ but I'm really losing it when it comes to indistinguishable objects.
I am  completely lost solving this. Can anyone help me understand the problem in a easier way so that I can solve it?
P.S. : I don't need full solution, just some help understanding it, hints at max so that I can solve it on my own.
 A: $\mathbf{\text{Note 2}}=$ I am sorry, i did not see that you do not want the answers , so i have erased them now , if you find the solution , you can write them to comment section , and i can check them for you.
a-) It is said that the ipads are identical and the classrooms are distinct , so by using stars and bars there are ... ways , because there is not anymore restriction.
Lets come to the logic behind this method assume that the ipads are represented by "*" and $|$ represents the classrooms.If there were $5$ identical ipads and $3$ different classrooms then ,one of the distribution will be "$**|*|**$" ,by using combinantion with repetition ,we find the all arrangements of stars and bars formula ,$C(m+n-1 ,n)$ where m is the number of distinct boxes (in this question it represents classrooms , to satisfy the number bars ,we subtract $1$ from $m$)  and $n$ is the number of stars, you can also think their arrangements as permutation with repetition , so it give use all arrangements.In that example , we applied it for $60$ stars and $6$ bars , we used $6$ bars because it gives us $7$ places to represent $7$ classrooms.
b-) Select $4$ classroom that take ipads by $C(7,4)$ and again stars and bars ... , then $$C(7,4) \times ..$$
For more information as to star and bars , look at : https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
$\mathbf{\text{Note}}=$ If the ipads were distinct then , it would $\mathbf{\color{red}{\text{not}}}$ be equal to $60 \times 59 \times 58 ... 54$ as you said (if there is not restriction) , in that cases , you had better use exponential generating functions.
