Not a popular question - was it harder than I thought maybe?
Pen Areecharoenlert, Suzanne Abbott and Rachel Walker of The Mount School, York were close to the first part, showing there are only two essentially different "standard dice" with alternative faces adding to 7. To quote them,
". . . if you look at the corner surrounded by 1,2 & 3 you either count clockwise or anticlockwise . . .".
They used diagrams of the net of a cube to help them.
My solution is as follows;
so there are 2 and only 2 standard dice.
Suppose we relax the condition that opposite faces sum to 7. How many different dice can we make now? Pen, Suzanne and Rachel mention a school colleague in Year 12 who "wished to remain nameless" but who told them there were 6! = 6x5x4x3x2x1 = 720 different dice. They thought it ". . . seemed a lot . . .". Quite right, there are in fact only 30 possible dice.
I give two proofs for you to choose from;
so the number of choices is 5 x3 x 2 = 30
Write [(1,6), (2,5), (3,4)] to represent the "pairing" for the standard die.
so there are 30 dice altogether.
I stole the idea for the second proof from Chiara Colli's solution to the Russian Cubes problem of last month. Chiara is from the Liceo Cairoli in Vigevano, Italy.