Badinage, you are right.
There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd That is 28*27*26
BUT it doesn't matter what order they are chosen. so you have to divide by 3! = 6
$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$
this is how I would normally do it
28C3
$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$
Yes, that's the answer your teacher is seeking.
However, as a challenge extension you could extend this to include ties, e.g., from 2 up to 28 ties for first, up to 27 ties for second, etc. I'm not sure how to do this, though, and it sounds complicated.
.
Badinage, you are right.
There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd That is 28*27*26
BUT it doesn't matter what order they are chosen. so you have to divide by 3! = 6
$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$
this is how I would normally do it
28C3
$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$