Start with not using 478
then you have 2 x 2 x 2 x 2 x 2 = 32 combos
then you can have ONE 8 or ONE 7 or ONE 4 each in one of the 5 positions = 15 the other 4 positions 2 x 2 x 2 x 2 possible
15 x 16 = 240
then you can have TWO of the 4 7 8 47 74 78 87 48 84 5 C 2 x 6 = 60
the other three positions would be 2 x 2 x 2 = 8
60 x 8 = 480
you can have all of 4 7 8 in 6 orders x 5 C3 = 60 the other two positions filled by 2 x 2
4 x 60 = 240
Summing the reds = 992 possible numbers ( same answer guest found !)