If the least two are on the same face, then the other two digits has to be larger than 9 – 1 – 2 = 6, so both { 7 and 8 } will be out of commission for other faces, however, if 1 and 2 are on the same edge as neighboring vertices, then both faces that shared said edge would need a total of 2 × (4 – 2) = 4 largest digits out of the two available, so this is impossible Answer = [ 4 × (4 + 1)/2 ] ÷ 2 = 5