Randomly dropping coins in columns, odds of all columns being equal? 
You have $n$ vertical tubes and you drop $c$ coins randomly in them.
  What are the odds that all tubes will contain equal amount of coins
  after $c$ drops?

Trivially, $c$ needs to be divisible by $n$, otherwise $P(c,n)=0$.

Looking at $n=2$, 
I draw a branching tree with states calculating the probability of each state, I noticed a pattern, which seems to be telling that for $c=2k$, we should have: $$  P(2,c)=\binom{2k}{k}\times\frac{1}{4^k}$$
Now to calculate $n=3,4,5\dots$ I would need to draw branching trees for each case, and hope to find a pattern to find a general expression.

Can we fully solve this more elegantly and quicker, without needing to draw probability trees and
  observe the patterns there?

I didn't have any lessons on probability or statistics in my high school curriculum, so I've always relied on inventing my own ways to solve problems like this one. 
 A: Here is a way to do it.
As you know, the probability of an event $E$ happening is equal to the number of possibilities in which it occurs divided by the total number of possibilities, assuming that all outcomes have the same likelihood.
Of course, if we have $c$ coins and $n$ tubes, then the total number of ways to distribute the coins is $n^c$.
Now we need to find then number of ways to partition $c$ coins into partitions of size $\frac{c}{n}$. Since you have stated that you have not taken any classes regarding this, you may not be aware of the "partition formula", which states that the number of ways of partitioning a group of $a$ things into groups of $b_1, b_2,...,b_k$ is given by
$$\frac{a!}{b_1!b_2!...b_k!}$$
Thus the number of ways to partition $c$ coins into groups of $(\frac{c}{n})$ is
$$\frac{c!}{(\frac{c}{n})!^n}$$
and so the probability is
$$\frac{c!}{n^c(\frac{c}{n})!^n}$$
if $n$ evenly divides $c$. Otherwise, as you have stated, the probability is $0$.
A: $$
P(n,nk)=\binom{kn}{k,\ldots,k}\frac{1}{n^{nk}},
$$
otherwise $P(n,c)=0$.
A: Use multinomial distribution to conclude that,
$$P(n,c) = \frac{(nk)!}{(k!)^n} \left(\frac{1}{n}\right)^{nk}$$
where $c = nk$
