Congratulations to Alex Summers, who successfully worked out the solution to our December maths challenge!

Alex noted correctly that Santa could team his reindeer in 5472 different ways. See below for his solution and thank you to everyone who took part. Don’t forget to take a look at January’s maths challenge.

Solution:

I consider choosing positions for the reindeer in the order: Donner + Blitzen, Vixen, Comet, all of the rest.

Donner and Blitzen have to be adjacent to each other:

CASE 1: Donner and Blitzen are in the same column.

Either can be in front: 2 different choices.

The front of this “pair” of reindeer can be positioned in any one of (4n-2) positions (not the back two positions).

Now we have two sub-cases:

CASE 1a: Vixen is also in the same column.

There are (2n-2) choices for Vixen’s position.

Since Comet must be in the same column, there are (2n-3) choices for Comet.

The remaining (4n-4) reindeer are unconstrained, so there are (4n-4)! choices for them.

So, for this (sub)case we have 2(4n-2)(2n-2)(2n-3)((4n-4)!) choices [in the simple version of the problem: 2x6x2x1x4!]

CASE 1b: Vixen is in a different column.

There are (2n-1) possible positions (cannot be next to Donner).

Comet must be in the same column as Vixen, for which there are also (2n-1) leftover positions.

The remaining (4n-4) reindeer are unconstrained, so there are (4n-4)! choices for them.

So, for this (sub)case we have 2(4n-2)(2n-1)(2n-1)((4n-4)!) choices [in the simple version of the problem: 2x6x3x3x4!]

CASE 2: Donner and Blitzen are in the same row.

Either can be on the left: 2 different choices.

There are 2n rows to choose from.

Vixen can be placed in any of the remaining (4n-2) positions.

Comet must go in the same column, for which there are now (2n-2) choices.

The remaining (4n-4) reindeer are unconstrained, so there are (4n-4)! choices for them.

So, for this case we have 2(2n)(4n-2)(2n-2)((4n-4)!) choices [in the simple version of the problem: 2x4x6x2x4!

In total, there are ((2n-2)(2n-3) + (2n-1)(2n-1) + (2n)(2n-2))(2(4n-2)((4n-4)!)) possibilities.

Or, in the simple version, (2 + 9 + 8)(2x6x4!) = 5472