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.

See OEIS sequence A001399. You are asking for the value of $a_{200}$ where $a_n$ is the number of partitions of the number $n$ into at most $3$ parts, or equivalently, the number of partitions of $n$ into parts of size at most $3$, that is, the number of ways we can write $n$ as a sum of $1$s, $2$s, and $3$s without regard to order.
Let $A_{n,i}$ be the set of all partitions of $n$ into parts of size at most $3$ with at least one part equal to $i$. By the in-and-out formula, for $n\gt0$ we have
$$a_n=|A_{n,1}\cup A_{n,2}\cup A_{n,3}|$$
$$=|A_{n,1}|+|A_{n,2}|+|A_{n,3}|-|A_{n,1}\cap A_{n,2}|-|A_{n,1}\cap A_{n,3}|-|A_{n,2}\cap A_{n,3}|+|A_{n,1}\cap A_{n,2}\cap A_{n,3}|$$
$$=a_{n-1}+a_{n-2}+a_{n-3}-a_{n-3}-a_{n-4}-a_{n-5}+a_{n-6}$$$$=a_{n-1}+a_{n-2}-a_{n-4}-a_{n-5}+a_{n-6}.$$
Now, the homogeneous linear recurrence
$$a_n=a_{n-1}+a_{n-2}-a_{n-4}-a_{n-5}+a_{n-6}$$
has the characteristic polynomial
$$t^6-t^5-t^4+t^2+t-1=(t-1)^3(t+1)(t^2+t+1)$$
with roots
$$1,\ 1,\ 1,\ -1,\ e^{2\pi i/3},\ e^{-2\pi i/3};$$
so the general solution is
$$a_n=An^2+Bn+C+D(-1)^n+E\cos\frac{2n\pi}3+F\sin\frac{2n\pi}3$$
where $A,B,C,D,E,F$ are arbitrary constants.
We use the initial values $a_0=1$ and $a_n=n$ for $n=1,2,3,4,5$ (or $a_n=0$ for $n=-1,-2,-3,-4,-5$) to evaluate the constants and get the particular solution
$$a_n=\frac{6n^2+36n+47+9(-1)^n+16\cos\frac{2n\pi}3}{72}.$$
Finally,
$$a_{200}=\frac{240000+7200+47+9-8}{72}=\boxed{3434}.$$
By the way, since
$$a_n=\frac{(n+2)(n+4)}{12}+\frac{-1+9(-1)^n+16\cos\frac{2n\pi}3}{72}$$
and since
$$\left|\frac{-1+9(-1)^n+16\cos\frac{2n\pi}3}{72}\right|\lt\frac12,$$
it follows that $a_n$ is the nearest integer to $\frac{(n+2)(n+4)}{12}$. For $n=200$ we have
$$\frac{(n+2)(n+4)}{12}=\frac{202\cdot204}{12}=3434=a_{200}.$$
