m-nomial coefficient While the binomial coefficient are the coefficients of \$(1+x)^n\$, m-nomial coefficients are the coefficients of \$(1+x+x^2+...+x^{m-1})^n\$.
For example, \$m(3,5,6)\$ is the coefficient of \$x^6\$ in the expansion of \$(1+x+x^2)^5\$.
Write a program/function that takes 3 numbers as input and outputs the corresponding m-nomial coefficient.
Details

Input can be taken via any reasonable method.
Everything out of range is \$0\$, for example \$m(2,0,-1) = m(2,5,6) = 0\$
The first argument will be positive.
The second argument will not be negative.
All three arguments will be integers.

Testcases
Inputs:
[1,0,0]
[2,5,6]
[3,5,6]
[4,5,-1]

Outputs:
1
0
45
0

Scoring
This is code-golf. Shortest solution in bytes wins.
Perl, 63 60 59 57 bytes

Includes +2 for -pa

Give input parameters on STDIN

mnomial.pl &lt;&lt;&lt; "3 5 6"

#!/usr/bin/perl -pa
$_=/-/^($_--,1x$F[2])=~/^(1{0,$_}){$F[1]}$(??{++$;})/+$


As a perl programmer I just know that every problem is really a regex in disguise...

Explanation:

m(3,5,6) can be seen as the number of ways you can get 6 by adding 5 integers in the range [0..2]. You can write a regex matching such a partition using:

1 x 6 = /^(1{0,2}){5}$/


However once a single solution is found the regex stops. We need to fail the found solution to get the regex to backtrack to the next solution and count the failures. In perl this can be done by extending the regex at runtime with something that is sure to fail, e.g. by demanding that something comes after the end of the string:

 1 x 6 = /^(1{0,2}){5}$(??{++$n})/


$n will count the number of failed attempts

My solution is basically this with two modifications:


Output the counter in such a way that you get 0 instead of the empty string if there is no valid partition and the counter never gets incremented
If the third argument is negative it would work if negative string lengths existed, but 1 x -5 is the empty string and 0 can be written exactly one way as sum of zeros, so that would output 1 instead of 0. I xor with the result of /-/to change the 1 back to 0


Without the need to handle a negative third argument this 49 byte solution would be good enough (give one parameter of m(x,y,z) on each line of STDIN in order y \n z \n x \n):

#!/usr/bin/perl -pa
$_=(1x&lt;&gt;)=~/^(1{0,@{[&lt;&gt;-1]}}){@F}$(??{++$;})/+$