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Please note: this competition is now CLOSED.

An apparently straightforward one this time:

In how many different ways can the numbers 1 to 6 on a die (cube) be marked? As usual, the numbers on opposite faces must add to 7.

Give your reasoning.

the entries

Many thanks to all of you who entered the brainbuster.

This one attracted a large number of entries once again.

We did say it was apparently straightforward - the answer depends on what is meant by 'different' (or equivalently, what is meant by the same) in this context. Also, whether the 'numbers' were digits or the usual dot patterns, and whether their orientation matters; whether the faces themselves are considered identical or different in some way...

There was a very wide range of answers given! Here's the range:

1
2
3! **
4
8
16
30
48
384
720
2048
8192
196608
infinity

** that's 3 followed by an exclamation mark to indicate amazement at the answer, not 3 factorial

Of these answers 2, 48 and (to a lesser extent) 16 were the most popular, with reasonably convincing explanations given for most replies.

Into the trusty ecm hat went all answers that flagged up the ambiguities and gave at least some of the possibile answers.

the solution

Here is the answer as provided by our winner:

There are (at least!) four distinct problems here, depending on whether the sides of the die are to be considered distinguishable or not i.e. whether solutions that are identical up to rotating the die are considered different, and whether the numbers are allowed to appear rotated once it has been dedided which face to put them on.

Firstly suppose that all faces of the die are coloured in a different manner i.e. solutions are not considered identical up to rotation.

Then there are 6 choices for '1', each leading to a distinct solution and each such choice forces '6' opposite. There are 4 reamining choices for '2' and each such choice forces '5' opposite. There remain two choices for '3' and '4' takes the remaining face.

Therefore in this context, there are 6 x 4 x 2 = 48 solutions.

Secondly, suppose that the faces of the die are identical and that we wish to consider the family of solutions that are equivalent under rotation of the die as identical.

Then we can place '1' anywhere without loss of generality since we are only interested in where the numbers lie in relation to each other. Then '6' must be opposite '1', and we can place '2' anywhere without any loss of generality, since wherever we place '2', simply rotating the die will get it to any of the other available places. This leaves two choices for where to place '3' and 4; and each choice gives a distinct solution. Thus in this case there are two distinct solutions.

Thirdly, note that the numbers themselves could be arranged in different manners, though I presume this was probably not intended as part of the solution as for example an upside down '6' could be interpreted as '9', though one could argue that as '9' is known not to be on the die, this is irrelevant.

In this case, firstly, if we treat the sides as distinct, we can fix the orientation of '1', and then for each of the 48 solutions, there are 4 different ways in which we can place each of the remaining numbers on the die (since none of the numbers two to six has a non-trivial rotational symmetry). Thus in this case, there are 48 x 4^5 = 49152 solutions.

Finally, if we regard solutions that are obtained from each other by a rotation of the die as equivalent, we have 2 x 4^5 = 2048 solutions. (Fix '1' for each of the 2 choices of where to place the numbers, then each of the other five numbers has 4 possible alignments).

Note that I have assumed the numbers will be placed centrally on the faces and aligned so that they are parallel to the sides. Without these assumptions, there are infinitely many solutions!

and the lucky winner is...

Congratulations to Dominic Goodwin, a PhD-qualified Software Developer, who wins £200 of Firebox vouchers.

Many thanks again to all entrants - do try the next brainbuster!