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For an exercise in my math class we are calculating the cost of the 12 days of christmas.

Let's define a set $c$ to be the price of each item in the popular "12 days of christmas" song, from a partridge in a pear tree all the way up to the drummers drumming.

From there I have figured out the formula for the total cost is this:

$$\sum_{n=1}^{12}c_n(n (13-n))$$

And this works. But, it isn't exactly something that you can calculate easily on paper. Is there a way to calculate this cost without having to manually calculate each one and then add it up, perhaps even a way to do it in your head?

EDIT: For those not familiar with the song, here is the lyrics, and the list of items is as follows:

1 Partridge in a pear tree
2 Turtle Doves
3 French Hens
4 Calling Birds
5 Gold Rings
6 Geese a-Laying
7 Swans a-Swimming
8 Maids a-Milking
9 Ladies Dancing
10 Lords a-Leaping
11 Pipers Piping
12 Drummers Drumming
Nico A
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  • Can you please list the items in that song for those of us who are not acquainted with it? – barak manos Dec 16 '15 at 12:16
  • There's an old Peanuts comic strip in which Linus enumerates how many items of each kind there are (i.e., 12 partridges in a pear tree, 22 turtle doves, 30 French hens, etc.), but I don't think he adds up the total item count. I wonder if someone can find it. – Brian Tung Dec 16 '15 at 18:38

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I think that, unless we are told anything about the relationship between the costs of a drumming drummer, a piping piper, a leaping lord, a dancing lady, a milking maid, a swimming swan, a laying goose, a golden ring, a calling bird, a french hen, a turtle dove and a partridge, then there's not much simplification to be done.

Arthur
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Arthur is right - the relationship between $c_n$ is required in order to simplify further.

If, instead, $c_n=1$ (all objects cost $1), the total price is

$$\sum_{n=1}^{12}n(13-n)=\sum_{n=1}^{12}\sum_{m=n}^{12}n=\sum_{m=1}^{12}\sum_{n=1}^m \binom n1=\sum_{m=1}^{12}\binom {m+1}2=\binom{14}3=364\quad\blacksquare$$

This also means that the total number of objects is $364$.

Curiously and coincidentally, this also means that there is one gift for each day of the year*, apart from one day - perhaps Christmas Eve!

*assuming a non-leap year

Community
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Hypergeometricx
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  • I will add that it is possible to find some posts about this sum for $c_n=1$. In fact, the answer given here mentioned exactly this song. Some other related posts can be found among questions which are linked there. – Martin Sleziak Dec 16 '15 at 14:09
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When all $c_n=1,$ we are looking at diagonals in Pascal's triangle, as in the answer by hypergeometric

enter image description here

Will Jagy
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    And I don't think it's a coincidence that this looks like a Christmas tree! – pjs36 Dec 16 '15 at 18:29
  • In fact the gifts can be arranged in tetrahedron with the $n$th layer containing gifts from the $n$th day, resulting in a nice tetrahedral Christmas tree! The total number of elements is a tetrahedral number. – Hypergeometricx Dec 17 '15 at 16:41
  • @hypergeometric, yes. In American pool, related to billiards, the "rack" is a plastic triangle into which fit 15 balls in an equilateral triangle. One may pile more layers and get a tetrahedron of 35 balls. – Will Jagy Dec 17 '15 at 17:45
  • You threw up Pascal's triangle without explaining how to use it here. – moonman239 Dec 24 '21 at 21:47