Compute the Eulerian number The Eulerian number A(n, m) is the number of permutations of [1, 2, ..., n] in which exactly m elements are greater than the previous element. These are also called rises. For example, if n = 3, there are 3! = 6 permutations of [1, 2, 3]

1 2 3
 &lt; &lt;  2 elements are greater than the previous

1 3 2
 &lt; &gt;  1 ...

2 1 3
 &gt; &lt;  1 ...

2 3 1
 &lt; &gt;  1 ...

3 1 2
 &gt; &lt;  1 ...

3 2 1
 &gt; &gt;  0 ...


So the outputs for A(3, m) for m in [0, 1, 2, 3] will be

A(3, 0) = 1
A(3, 1) = 4
A(3, 2) = 1
A(3, 3) = 0


Also, this is the OEIS sequence A173018.

Rules


This is code-golf so the shortest code wins.
The input n will be a nonnegative integer and m will be a integer in the range [0, 1, ..., n].


Test Cases

n   m   A(n, m)
0   0   1
1   0   1
1   1   0
2   0   1
2   1   1
2   2   0
3   0   1
3   1   4
3   2   1
3   3   0
4   0   1
4   1   11
4   2   11
4   3   1
4   4   0
5   1   26
7   4   1191
9   5   88234
10  5   1310354
10  7   47840
10  10  0
12  2   478271
15  6   311387598411
17  1   131054
20  16  1026509354985
42  42  0
Jelly, 8 bytes

Œ!Z&gt;2\Sċ


Try it online! (takes a while) or verify the smaller test cases.

How it works

Œ!Z&gt;2\Sċ  Main link. Arguments: n, m

Œ!        Generate the matrix of all permutations of [1, ..., n].
  Z       Zip/transpose, placing the permutations in the columns.
   &gt;2\    Compare columns pairwise with vectorizing greater-than.
          This generates a 1 in the column for each rise in that permutation.
      S   Compute the vectorizing sum of the columns, counting the number of rises.
       ċ  Count how many times m appears in the computed counts.