Is it possible to emplace a std::array in a container? If it is how? If not, why? [duplicate]

Given the simple code

#include <array>
#include <vector>
int main() {
    std::vector<std::array<int,3>> v;
    v.emplace_back(std::array<int,3>{1,2,3});
}

I'm first of all worried about what is really going on.

My understanding of emplace_back is that it constructs an element (in this case an std::array<int,3> with values 1, 2, 3) in-place by forwarding its argument(s) to that element's constructor.

But which constructor? Since we are passing an object of the same type of the element we want to emplace, I guess the move constructor is selected, as we're passing a temporary, or the copy ctor if the move ctor is deleted (implicitly or explicitly). Besides, we are first of all constructing that temporary, and that's not done in place, right?

So we are basically calling the "normal" (???) constructor of std::array<int,3> by passing to it 1, 2 , and 3 as arguments, thus getting back a temporary, which is passed to the copy ctor of std::array<int,3> which constructs the copy in place or, preferably, to the move ctor of std::array<int,3> which (?) copies the underlying C-style pointer?

It should be clear this simple example is confusing me a lot.

To confuse me even more is the doubt that if I really wanted to take advantage of emplace_back, I could pass to it only the arguments to the std::array<int,3>, as in

#include <array>
#include <vector>
int main() {
    std::vector<std::array<int,3>> v;
    v.emplace_back(1,2,3);
}

which doesn't even compile.