Suppose we have $n$ points. Pick $k$ of them with replacement. Assume the order does not matter. How can we do this?
Pick the first element, n ways to do this. since we dont replace same happens for second element and so forth. In total we must have $n^k$ ways. My notes say the correct answer is ${ n + k - 1 \choose k }$ which I find mind blowing how this is possible.
I will assume that the objects are distinguishable, each being of a certain unique type. Your answer would be correct if the order did matter but it does not. Therefore two ways to pick $k$ objects are distinguished only by the numbers of picked objects of each type.
There are altogether $n$ types, the total of chosen objects being $k$. Thus, the problem is reduced to the question in how many ways $k$ can be partitioned into $n$ non-negative parts. This is classical stars and bars problem with $k$ stars and $n-1$ bars, and the answer is indeed: $$ \binom{k+n-1}{k}. $$