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Algebra
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I guess there is a simpler solution . Let us look at any combination so that we always face the side painted with the same color (say, red). In this case we have 5 sides left, therefore there is 5! = 120 ways to paint other 5 sides.
Now the uniqueness comes into play. Still holding the cube facing the selected (red) side, just rotating it around the axis, perpendicular to this selected side, we can put it in 4 different orientations, obviously repeating each other. THerefore we have to consider only one out of orientations, and the number of solutions is 5!/4 = 120/4=30
Even i get it as 30....
say colours are P Q R S T U
let ny sides 1 & 2 (opp. sides) be painted with colours P & Q
so v are left with 4 sides n 4 colours so ways = 4!
but notice tht v can rotate it either way 2 get similar orientations...
like RSTU ...STUR.... TURS ....RSTU (going in order) wud mean the same..
so ways fr the 4 sides = 4!/4 = 3! (v cud also hav considered circular PnC directly fr this!)
now if v keep ny one fixed (eg P on side 1) side 2 can hav "5" choices....according 2 which rest of the 4 colours wud b arranged...
So total ways = 5*3! = 30 ways...













6!*6!+5!/2 Is it correct. Verify