3/7 of the cards are even, they need to be on the end. (3/7)1260 = 540
this time the last two digits need to be a multiple of 4 - either 16,16,24,32,32. there are 7C2 = 21 different combinations of the last two numbers, so (5/21)1260 = 300
again, 21 combinations, this time only 1-1 and 3-3 are possibilities, so (2/21)1260 = 120