An imperial method. Nowhere near as fancy.
Just arrange the permutations (removing repetitions) of
0, 0, 0, 6                = 4!/3! = 4
0, 0, 1, 5                = 4!/2! = 12
0, 0, 2, 4                = 4!/2! = 12
0, 0, 3, 3                = 4!/(2!*2!) = 6
0, 1, 1, 4                = 4!/2! = 12
0, 1, 2, 3                = 4! = 24
0, 2, 2, 2                = 4!/3! = 4
1, 1, 1, 3                = 4!/3! = 4
1, 1, 2, 2                = 4!/(2!*2!) = 6

which adding up gives 84 the required number.