Put identical coins into identical boxes In how many ways can I put 200 (identical) coins into 3 identical boxes?
I started count it, and I saw that the possibilities goes this way:
$200, 0, 0$
$199, 1, 0$
$198, 2, 0$
$198, 1, 1$ 
$197, 3, 0$
$197, 2, 1$
ect.
which mean I can count it this way: $2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3$ ect..
I'm wondering if there is a better way to count all the possibilities.

First label the boxes A,B, and C. Now we can find the number of tuples (x,y,z) of configurations in which there are x in A, y in B and z in C. We do this by a method commonly known as Stars And Bars 
The number of tuples happens to be $\binom{202}{2}$
Now to treat the boxes as identical, we must filter out the dupicates. First we remove the cases where there x=y. Then we have something of the form 2x+z=200. Observe that the solutions are (x,z)=(0,200)(1,198)(2,196)(3,194)(4,192)...(100,0)
It should be clear that there are 101 such ways. To include the cases where y=z and x=z, we multiply by three, so we exclude 303 tuples (as they have been double counted), divide by three factorial, then add 101 back later (to compensate).
Observe that in each of the above three cases, we have no possibility of over-counting as we could only over-count if x=y=z, which is impossible as 3 does not divide 200.
So now the number of tuples where x,y and z are distinct is $\binom{202}{2} - 303$
Now we account for all permutations of x,y and z to find the number of ways denoted [x,y,z]. We do this by dividing by three factorial. Note that $\binom{202}{2}-303$ is divisible by six.
Thus we find that the number of ways of distributing 200 identical balls in 3 identical boxes is $\frac{\binom{202}{2}-303}{3!}+101 = 3434$
