Number of ways to seat $n_i$ people around a round table with $n$ seats $\bigl(n_i \lt n \bigl)$

Question

So the question says to find the number of ways to seat $n_i$ people around a roundtable with $n$ seats $\bigl(n_i \lt n \bigl)$.
The first thing I did was select $n_i$ seats from $n \Rightarrow \binom {n}{n_i}$
Now the number of ways to arrange these $n_i$ people is obviously $\bigl(n_i -1 \bigl)!$
So we get, $$\binom {n}{n_i} \cdot \Bigl(n_i -1\Bigl)!$$ The answer however given is $$\left( \frac{n!}{\Bigl(n-n_i\Bigl)!}\right)$$ Please tell me where I am going wrong.

Answer 1

The answer is finding all permutations and dividing by the number of spaces since they are indistinguishable. However, you should note that there are $n_i!$ ways to arrange them since they can have any permutation. (Rotations are distinguishable in this case, since there is nothing to account for overcount in the answer.)

When we implement this, we get $\dfrac{n!}{n_i!(n-n_i)!}\cdot n_i!=\dfrac{n!}{(n-n_i)!},$ which is the answer given.