I was looking up the lyrics of *The Twelve Days of Christmas* song (you know, the one that goes “On the first day of Christmas, my true love sent to me…”) and Simon said he actually wanted to know how many presents the lucky singer got from her true love. Because she got one partridge in a pear tree on the first day, then two turtle doves and another partridge in a pear tree on the second day, then three French hens, another pair of turtle doves and yet another partridge in a pear tree on the third day, and then on the fourth day, she got four calling birds, and – you guessed it – three more French hens, two more doves and – why did he keep doing that to her? – another freaking partridge! In another pear tree! In the end, she must have had 12 partridges in 12 trees, 22 doves, 30 French hens and 36 calling birds. That’s a lot of bird poop! And the true love was only just getting started, that was all just a warm-up. Luckily, he stepped over from poultry to gold on day five, so the girl at least got some durable assets added in there: 40 golden rings (which is also strange as she couldn’t possibly wear the rings on her toes). But what did that bastard do next? He started throwing birds at her again! 42 geese a-laying (so more geese coming on top of our calculation!) and 42 swans a-swimming (I guess they came with an inflatable pool of some sort). By this time it was already one loud stinking mess, but it’s what he did next that was absolutely unacceptable: he started propping her home with hoards of strangers! He gave her 40 maids a-milking (did they come with a cow each?), 36 ladies dancing (was that a form of human trafficking?) and topped those with 30 lords a-leaping (I don’t even dare to picture those, especially since they didn’t get enough ladies). And just when she couldn’t take it anymore and probably tried begging him for a quantum of solace and quiet, he shove 22 pipers piping in there and 12 drummers! At that point they probably didn’t even hear each other anymore. It’s actually quite a nightmare of a story, if you think about it. Possibly even an early attempt of critique on consumerism.

But funnies aside, Simon came up with a beautiful geometrical visualization of how to calculate the total number of gifts that quickly yielded a formula (the tetrahedral numbers formula!), akin to the famous Gauss trick (Simon did the writing from here):

With the Gauss trick, when you have a sum like 1 + 2 + 3 + 4 + 5 + 6, you first write it down backwards, like 6 + 5 + 4 + 3 + 2 + 1. Then you add the two sums together, and then you get 7 + 7 + 7 + 7 + 7 + 7. That’s just 6 * 7! Though we do have to divide through by 2, because, remember, this was not the original sum. It was the original sum PLUS the original sum backwards. Anyway, this gives 6*7/2, which is 28. If you do this again, but in general, you get a formula of n(n+1)/2.

Now, how can we extend this to the 12 days of Christmas? It’s actually a bit more complicated than you think. For the 12 days of Christmas, it’s not just a sum of the numbers 1 through 12. No, on the first day you get just 1 present, a partridge. On the second day, you get 2 turtle doves and 1 partridge, which is 1 + 2. On the third day, you get 3 French hens, 2 turtle doves, and 1 partridge in a pear tree. That’s 1 + 2 + 3. And so on. So the actual sum we want to calculate is 1 plus 1 + 2 plus 1 + 2 + 3 plus 1 + 2 + 3 + 4 plus and so on.

Now we’re ready to apply the modified version of the Gauss trick. We can arrange the sum in a triangle:

Now, remember how we wrote the sum backwards with the normal Gauss trick? Well, here, we can rotate the triangle in 2 different ways:

3 triangles. We can add them together:

We get all 6s! How many 6s are there? Well, it’s a triangle of 6s, where the side length is the number that we started with, which in this case is 4. Applying the normal Gauss trick to that we get 10. So there are 10 6s! We can calculate that 10*6 is 60. This is not the answer, though. Remember in the normal Gauss trick when we had to divide by 2, because we added 2 copies of the sum? Well, here, we added 3 copies of the triangle, so we have to divide through by 3. This gives 10*6/3, which is 20.

We can do this in general, with n instead of 4. In that case, we get n(n+1)/2 (n+2)’s. So we get n(n+1)(n+2)/2. After dividing by 3, we get n(n+1)(n+2)/6. So that’s the general formula!

So, drumroll (that’s those 12 drummers still going!) and let’s plug the song’s numbers in: 12(12+1)(12+2)/6 equals… 364!

That’s like getting a present every day of the year. Except for one lucky day. Because, let’s face it, you would return most of those presents, wouldn’t you?

\* I hope no one notices the photoshops that we did on some of the images.

Just curious. You say “we get n(n+1)/2 (n+2)’s”. I can see that you get n(n+1)6 from the illustration provided but did you attempt to prove that the sum at each position in the triangle was n+2 in the general case?

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Simon replies: OK, take the n=4 case I used as an example in the post. Look at the top corners of the three triangles. Two of them are 1, and the other one is 4. So the value in the final triangle is 4+2, which is 6. You can use a similar argument to do this in general, and you get n+2 as a result.

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