Basically, yes. As it turns out, this is much easier than I expected. In c#, again (and using the same Choose() function in my earlier post:

Code:
private long[] CalculateAll()
{
    long[] total = new long[5];

    for(long i=0; i<=4;i++)
        for(long j=0; j<=4;j++)
            for(long k=0; k<=4;k++)
                for(long l=0; l<=4;l++)
                    total[Math.Min(Math.Min(i,j),Math.Min(k,l))]+= Choose(4,i) * Choose(4,j) * Choose(4,k) * Choose(4,l) * Choose(64, 20 - (i+j+k+l));

    return total;
}
That's it! This nets you:
[4] = 635,376
[3] = 1,299,449,210,880
[2] = 7,576,701,736,868,160
[1] = 731,135,833,306,109,440
[0] = 2,796,602,307,719,350,464
[T] = 3,535,316,142,212,174,320

([T] is the sum of [0] - [4], and, as verification, equals (80,20))