Number of necessary stickers to complete a sticker album I have the following problem, and I was hoping you guys could help me solve it:

Consider a set of $t$ unique, collectable stickers (that accounts for the universe of collectable stickers, i.e., any sticker you buy is one of the $t$ unique stickers. Obviously, you can get repeated stickers - stickers that have the same unique identifier). You start with an album that is initially empty. You bought $c$ stickers. Assume the equiprobability of getting any sticker among the universe of $t$ different stickers. What is the probability of being able to complete the sticker album with the $c$ stickers you bought?

I know that if $c < t$, then such probability is $0$. I first tried solving the problem in the following manner: Given $f(1), f(2), ..., f(t)$ the number of stickers of $c$ that correspond to each sticker in $t$ (i.e., the number of repeated stickers you got for each of the $t$ possibilities), the linear equation I tried solving is:
$$f(1)+f(2)+...+f(t)=c \qquad \qquad \qquad (1)$$
Assuming that this can be modelled to the problem of counting the number of solutions of $(1)$ that are $\geq1$, and dividing by the number of solutions of $(1)$ that are $\geq0$. I got $\dfrac{c!(c-1)!}{(c-t)!(c+t-1)!}$ as the answer for that, but soon realised that the problem cannot be thought of as I assumed. For instance, say $t=c=2$. Then $1+1=2$ as the solution for $(1)$ doesn't have the same probability as, say, $0+2=2$. Any ideas on how to solve this? Any help is very much appreciated! Thank you!
 A: You appear to be using the stars and bars method.
You effectively have $t$ bins to place $c$ picks.  You want the probability that at least one pick is in each bin. (The 'bins' being the identity of the sticker.)
There are ${t+c-1\choose t-1}$ ways to fill the bins in total and ${c-1\choose t-1}$ ways to fill then with at least one pick.  (See Wikipedia)
$$P=\frac{c-1\choose t-1}{t+c-1\choose t-1}=\frac{c!(c-1)!}{(t+c-1)!(c-t)!}$$
The reason this isn't working is that the picks are not indistinct; they are enumerated by order of occurrence.  'Stars and bars' requires indistinct objects to be placed in distinct bins.
A: Reposted from a comment that the OP described as answering the question.

This sounds like what’s often called the “coupon collector’s problem.” More commonly-asked questions than yours are how many stickers do you need to buy to have a particular chance of getting all the stickers/coupons, or how many stickers do you need to buy, on average (one at a time) to get all varieties, but I think if you search for “coupon collector’s problem” (MSE, Google) you’ll find some useful approaches to answering your question. 
