How to find unique multisets of n naturals of a given domain and their numbers?

Hint: If $O_i$ is the number of occurrences of $i$ in the multiset, the tuple $(O_1, O_2, \dots, O_n)$ uniquely identifies the multiset. What constraint can you write down with the $O_i$, if the multiset has exactly $k$ elements?

Ok, to elaborate.

The number of multisets of size $k$ is equal to the number of non-negative integral solutions to the equation

$$O_1 + O_2 + \dots + O_n = k$$

Since the stars and bars method was already mentioned, here is a derivation of the formula using a more general approach, which is one of the main building blocks of analytic number theory: Generating Functions.

The number of solutions to the above equation is same as the coefficient of $x^k$ in

$$(1 + x + x^2 + x^3 + \dots )(1+x + x^2 + x^3 + \dots) \dots (1+ x + x^2 + x^3 + \dots) \ \text{repeated} \ n \ \text{times} \ \text{(why?)}$$

The above is same as

$$ \frac{1}{(1-x)^n} = (1-x)^{-n}$$

as $$ (1 + x + x^3 + x^3 + \dots) = \frac{1}{1-x}$$

Now we apply binomial theorem, (which is also valid for negative exponents)

The coefficient of $x^k$ is $$-1^{k}\ \frac{-n(-n-1)(-n-2)\dots(-n-(k-1))}{k!} = \frac{n(n+1)(n+2)\dots(n+k-1)}{k!}$$

$$= \frac{(n-1)!n(n+1)(n+2)\dots(n+k-1)}{(n-1)!k!}= {n+k-1 \choose k} $$