Explanation (slightly outdated)

The algorithm is pretty simple. Let's assume there are \$n = 3\$ contestants and Alice's rank \$k = 1\$. First, it generates a list of possible rankings by taking the Cartesian product of \$[0, 3)\$ with itself \3ドル\$ times. 0-indexing the ranks because it's more convenient to work with.

[(0, 0, 0), (0, 0, 1), (0, 0, 2),
 (0, 1, 0), (0, 1, 1), (0, 1, 2),
 (0, 2, 0), (0, 2, 1), (0, 2, 2),
 (1, 0, 0), (1, 0, 1), (1, 0, 2),
 (1, 1, 0), (1, 1, 1), (1, 1, 2),
 (1, 2, 0), (1, 2, 1), (1, 2, 2),
 (2, 0, 0), (2, 0, 1), (2, 0, 2),
 (2, 1, 0), (2, 1, 1), (2, 1, 2),
 (2, 2, 0), (2, 2, 1), (2, 2, 2)]

Then, it selects the potential rankings where Alice's rank, or the first rank listed, is \1ドル\$. Of course, it's still 0-indexed, so the following lists out rankings that start with \0ドル\$ instead.

After that, it knocks out invalid rankings by checking if each rank is equal to "the number of other participants who performed strictly better than them", or in other words, the number of ranks lower than the rank in question.

Explanation

The algorithm is pretty simple. Let's assume there are \$n = 3\$ contestants and Alice's rank \$k = 1\$. The ranks that follow will be 0-indexed because it's more convenient to work with. First, it generates a list of possible rankings where Alice's rank, or the first rank in the list, is \0ドル\$. This is done by taking the Cartesian product of the singleton list \$[0]\$ with \$n - 1 = 2\$ copies of \$[0, 3)\$.

Then, it knocks out invalid rankings by checking if each rank is equal to "the number of other participants who performed strictly better than them", or in other words, the number of ranks lower than the rank in question.

Python 3, (削除) 124 (削除ここまで) 119 bytes

lambda n,k:sum(all(r==sum(q<r for q in s)for r in s)*(s[0]==k-1)for s in product(*[range(n)]*n))
from itertools import*

Try it online!

Explanation

Python 3, (削除) 124 (削除ここまで) (削除) 119 (削除ここまで) 117 bytes

lambda n,k:sum(all(r==sum(q<r for q in s)for r in s)for s in product([k-1],*[range(n)]*(n-1)))
from itertools import*

Try it online!

Explanation (slightly outdated)

Python 3, 124 bytes

lambda n,k:sum(all(r==sum(q<r for q in s)for r in s)and s[0]==k-1for s in product(range(n),repeat=n))
from itertools import*

Try it online!

Python 3, (削除) 124 (削除ここまで) 119 bytes

-5 bytes thanks to enzo .

lambda n,k:sum(all(r==sum(q<r for q in s)for r in s)*(s[0]==k-1)for s in product(*[range(n)]*n))
from itertools import*

Try it online!