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Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers.

How Many Dice?

Stage: 3 Challenge Level: Challenge Level:1

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;

  • write 1 on a face and hold it with the face numbered 1 downwards.
  • the number 6 must go on the top.
  • write 2 on any blank face and rotate it so the 2 is away from you.
  • the 5 must now go on the face nearest you.
  • this leaves 2 blank faces to the Left and to the Right.
  • you have 2 choices when writing on the 3 and 4.

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;

proof #1

  • write 1 on the bottom face - this gives 5 choices for the top.
  • write any unused number on a blank face and rotate so the new number is away from you - this gives 3 choices for the front.
  • the remaining 2 numbers can go left/right or right/left - 2 choices.

so the number of choices is 5 x3 x 2 = 30

proof #2

Write [(1,6), (2,5), (3,4)] to represent the "pairing" for the standard die.

  • there are 2 dice for any given pairing.
  • there are 5 numbers to choose from to pair with 1
  • choose an unused number - there are 3 possible pairs for this one.
  • the remaining 2 numbers must be paired.
  • so there are 15 possible pairings, each leading to 2 possible dice.

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.