Let d[i] be the number of ways we can split i*4 golfers into groups of 4 golfers each.

We have d[1] = 1 and d[i] = d[i - 1] * choose(i * 4, 4) / i, where choose(x, y) is the binomial coefficient, calculated using the formula: choose(x, y) = x! / y! / (x-y)!.

When going from d[i-1] to d[i], we can pick any 4 golfers to be in the newly formed group (choose(i * 4, 4)). However, we need to discount repetitions, thus need to divide by the number of groups i.

The final solution is d[N / 4]^R.

Perl solution (Marius Gavrilescu)

In languages with builtin big numbers, we can compute d[N / 4] directly. Since d[N / 4]^R has more than 200 digits, we will print d[N / 4]^R.

    1 #!/usr/bin/perl
    2 use v5.14;
    3 use warnings;
    4 use bigint;
    5 
    6 my ($n, $r) = split ' ', <>;
    7 my $ans = 1;
    8 $ans *= (4 * $_)->bnok(4) / $_ for 2 .. $n / 4;
    9 say "$ans^$r";

C solution (Marius Gavrilescu)

We know that choose(K,4)=K!/(K-4)!/4! = K*(K-1)*(K-2)*(K-3)/24.

By expanding the definition of d we get d[N/4] = N*(N-1)*(N-2)*(N-3) / 24 * (N-4)*(N-5)*(N-6)*(N-7) / 24 * ... * 4*3*2*1 / 24 / (N/4)! which is equal to N! / (N/4)! / 24^(N/4).

Thus, the answer is (N*(N-1)*...*(N/4+1)/24^K)^R.

    1 #include <stdio.h>
    2 
    3 int main(void) {
    4     int n, k;
    5     scanf("%d%d", &n, &k);
    6 
    7     printf("(%d", n);
    8     for(int i = n - 1 ; i > n / 4 ; i--)
    9         printf("*%d", i);
   10     printf("/24^%d)^%d\n", n / 4, k);
   11     return 0;
   12 }

C++ solution (Gabriel Inelus)

We denote the number of ways to split N golfers into groups of 4 each when the order is important by the function F(x) = x choose 4 * F(x-4). To avoid the repetitions we can divide the answer by (x/4)! and then raise it to the power R. It's easy to observe that x choose 4 = x*(x-1)*(x-2)*(x-3)/24, which is less than 2³¹, thus we can give a straight forward recursive implementation of F(N).

    1 #include <bits/stdc++.h>
    2 using namespace std;
    3 
    4 void F(int k){
    5     cout << k*(k-1)*(k-2)*(k-3) / 24;
    6     if(k == 4) return;
    7     cout << "*";
    8     F(k - 4);
    9 }
   10 
   11 int main()
   12 {
   13     int N,R;
   14     cin >> N >> R;
   15     cout << "(("; F(N); cout << ")";
   16     for(int i = 2; i <= N/4; ++i)
   17         cout << "/" << i;
   18     cout << ")^" << R;
   19     return 0;
   20 }