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The Twelve Days of Christmas - How Many Presents?

By Murray Bourne, 16 Dec 2008

Partridge in a pear tree
A partridge in a pear tree

Most people wrongly believe that the '12 Days of Christmas' refers to the days before Christmas. However, it's really the period starting on Christmas day and finishing with the Epiphany (January 6th, when the 3 kings from 'the East' brought gifts).

You may be familiar with the Christmas song, The 12 Days of Christmas. The first few lines go like this:

On the first day of Christmas,
my true love sent to me
A partridge in a pear tree.

On the second day of Christmas,
my true love sent to me
Two turtle doves,
And a partridge in a pear tree.

On the third day of Christmas,
my true love sent to me
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

The song continues, adding 4 calling birds on the 4th day, 5 golden rings on the 5th, and so on up to the 12th day, when 12 drummers add to the cacophony of assorted birds, pipers and lords leaping all over the place.

Notice that on each day there is one partridge (so I will have 12 partridges by the 12th day), and each day from the second day onwards there are 2 doves (so I will have 22 doves), and from the 3rd there are 3 hens (total of 30 hens), and so on.

So, how many presents are there altogether?

Partridges: 1 × 12 = 12

Doves: 2 × 11 = 22

Hens 3 × 10 = 30

Calling birds: 4 × 9 = 36

Golden rings: 5 × 8 = 40

Geese: 6 × 7 = 42

Swans: 7 × 6 = 42

Maids: 8 × 5 = 40

Ladies: 9 × 4 = 36

Lords: 10 × 3 = 30

Pipers: 11 × 2 = 22

Drummers: 12 × 1 = 12

Total = 364

We observe that we have the same number of partridges as drummers (12 of each); doves and pipers (22 of each); hens and lords (30 of each) and so on. So the easiest way to count our presents is to add up to the middle of the list and then double the result: (12 + 22 + 30 + 36 + 40 + 42) × 2 = 364.

Efficiently counting the presents

What if we have more than 12 days? How many presents then?

Let's now generalize the above result just in case out true love decides to be extraordinarily generous and keeps on giving us gifts - up to 100 days, say. (Multiplying and adding could get quite tedious.)

Mathematically speaking, my true love is giving me 1 + 2 + 3 + ... + n presents on the n-th day after Christmas.

The number of presents each day is 1 on the 1st, then 3 on the 2nd, then 6 on the 3rd, then 10 on the 4th. We call this set of numbers the triangular numbers, because they can be drawn in a dot pattern that forms triangles:

Triangular numbers

To get the total number of presents, we need to add those triangular numbers, like this:

1 (on the first day) + 3 (on the 2nd day) + 6 + 10 + ...

Another way of writing this is:

On the first day, 1 present.
On the 2nd day, 1 + 3 = 4 presents
On the 3rd day, 1 + 3 + 6 = 10 presents
On the 4th day, 1 + 3 + 6 + 10 = 20 presents.

These partial sums are called tetrahedral numbers, because they can be drawn as 3-dimensional triangular pyramids (tetrahedrons) like this:

Tetrahedrons

So how many dots (representing presents) will there be in the 12th tetrahedron?

Of course, we could just start adding with our calculator, but what if my true love is very generous, and starts giving me presents for 30 days after Christmas? Or for 100 days? How would I calculate it then?

Our aim is to produce a formula that will allow us to find any tetrahedral number. Here's one of the possible ways of doing this.

Let's take (for example) the sum of the first 4 triangular numbers and represent it as a triangle. Each row in the triangle (on the left, below) adds to a triangular number and the sum of the whole triangle is the sum of the first 4 triangular numbers. Let's now re-arrange the first triangle in 2 different ways, then add the result, in respective positions. (My total is 3 times what I really need. I will divide by 3 later to cater for this).

Finding the n-th tetrahedral number

Notice that by doing this, I get a total of 6 in every position in the result triangle. The answer of "6" is 2 more than the 4 triangular numbers that we are adding. So if we were adding the first 7 triangular numbers, our result in the right triangle would be all 9's; if it was the first n triangular numbers, we would get (n + 2).

It's easy to find the sum of the 6's, like this:

(1 + 2 + 3 + 4) × 6.

The series in brackets is just an arithmetic progression, with first term a = 1, common difference d = 1 and number of terms n = 4.

The formula for the sum to n terms of an arithmetic progression with first term a and common difference d is:

\text{Sum}={\frac{{n}}{{2}}}{\left[{2}{a}+{\left({n}-{1}\right)}{d}\right]}

Substituting our values, we have:

\text{Sum}={\frac{{4}}{{2}}}{\left[{2}{\left({1}\right)}+{\left({4}-{1}\right)}{\left({1}\right)}\right]}

= 2 × 5

= 10

So (1 + 2 + 3 + 4) × 6 = 2(5) × 6 = 60

But remember, this is 3 times what we really want, so the 4th tetrahedral number is

60/3 = 20

General formula

In general, for the sum 1 + 2 + 3 + ... + n:

\text{Sum}={\frac{{n}}{{2}}}{\left[{2}+{\left({n}-{1}\right)}\right]}

which is the same as

\text{Sum}={\frac{{n}}{{2}}}{\left({n}+{1}\right)}

Multiplying by the (n + 2) that we get from what I called 'the result triangle' earlier:

{\frac{{n}}{{2}}}{\left({n}+{1}\right)}{\left({n}+{2}\right)}

Dividing this by 3 (since we used 3 equivalent sum triangles to get this far) gives us the n-th tetrahedral number:

{\frac{{n}}{{6}}}{\left({n}+{1}\right)}{\left({n}+{2}\right)}

Back to our Christmas song.

On the 12th day, the number of presents will be

{\frac{{12}}{{6}}}{\left({13}\right)}{\left({14}\right)}={364}

If my true love gave me the presents in this pattern for 30 days, I would have this number of presents:

{\frac{{30}}{{6}}}{\left({31}\right)}{\left({32}\right)}={4960}

If it was 100 days, I would have this number of presents:

{\frac{{100}}{{6}}}{\left({101}\right)}{\left({102}\right)}={171},{700}

Good deal.

Merry Christmas, everyone.

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