And a Partridge, une Perdrix

Probability Level 2

"On the first day of Christmas, my true love gave to me / a partridge in a pear tree."

"On the second day of Christmas, my true love gave to / two turtle doves and a partridge in a pear tree."

$\vdots$

"On the twelfth day of Christmas, my true love gave to me

12 drummers drumming
11 pipers piping
$\ \ \ \ \vdots$
2 turtle doves
1 partridge in a pear tree."

How many gifts does my true love give to me in these twelve days of Christmas?

The answer is 364.

3 solutions

Arjen Vreugdenhil
Dec 10, 2015

On day $n$ , my true love gives me $1 + \cdots + n = \left(\begin{array}{cc} n+1 \\ 2\end{array}\right)$ gifts. In twelve days, it is $\left(\begin{array}{cc} 2 \\ 2\end{array}\right) + \left(\begin{array}{cc} 3 \\ 2\end{array}\right) + \cdots + \left(\begin{array}{cc} 13 \\ 2\end{array}\right) = \left(\begin{array}{cc} 14 \\ 3\end{array}\right)$ gifts. (See hockey stick identity .) This evaluates as $\left(\begin{array}{cc} 14 \\ 3\end{array}\right) = \frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1} = 14\cdot 13\cdot 2 = \boxed{364}$ gifts-- one for every day of the year except Christmas itself.

Alternative solution:

Every gift can be described by the number of the day $d$ , the number of the gift $g$ (with $g = 1$ for a partridge and $g = 12$ for a drummer drumming), and the sequential number $n$ of the gift on that day. Clearly, $1 \leq n \leq g \leq d \leq 12$ ; and any triple of numbers $n, g, d$ that satisfies this equation corresponds to a gift.

Every gift can therefore be described uniquely by three different numbers $1 \leq n < g+1 < d+2 \leq 14$ . The total number of gifts is equal to the number of ways in which three distinct numbers $(n, g+1, d+2)$ may be chosen from $1, \dots, 14$ . This is the number of combinations of 3 out of 14, $\left(\begin{array}{cc} 14 \\ 3\end{array}\right)$ .

For $n$ days of Christmas one could also calculate it as

$\displaystyle\sum_{k=1}^{n} k((n + 1) - k) = (n + 1)*\sum_{k=1}^{n}k - \sum_{k=1}^{n} k^{2} = (n + 1)*\dfrac{n(n + 1)}{2} - \dfrac{n(n + 1)(2n + 1)}{6} =$

$\dfrac{n(n + 1)(3(n + 1) - (2n + 1)}{6} = \dfrac{n(n + 1)(n + 2)}{6},$

which is equivalent to $\dfrac{(n + 2)!}{((n + 2) - 3)!*3!} = \dbinom{n + 2}{3},$ as you have already ascertained.

Then for $n = 12$ there would be $\dfrac{12*13*14}{6} = 2*182 = \boxed{364}$ gifts. It does seem quite a coincidence that there ends up being one gift for every day besides Christmas, (except in a leap year). I wonder if the song writer "did the math" beforehand? :)

Brian Charlesworth - 5 years, 6 months ago

Yes-- it appears that you separate the sum into the different kinds of gifts; gift $\#k$ occurs on $n+1-k$ days, $k$ times each, hence the term $k((n+1)-k)$ .

There are

$1 \times 12 = 12$ partridges
$2 \times 11 = 22$ turtle doves
$3 \times 10 = 30$ French hens
$4 \times 9 = 36$ calling birds
$5 \times 8 = 40$ gold rings
$6 \times 7 = 42$ geese-a-laying
$7 \times 6 = 42$ swans-a-swimming
$8 \times 5 = 40$ maids milking
$9 \times 4 = 36$ ladies dancing
$10 \times 3 = 30$ lords-a-leaping
$11 \times 2 = 22$ pipers piping
$12 \times 1 = 12$ drummers drumming

Arjen Vreugdenhil - 5 years, 6 months ago

@Brian Charlesworth Sir, wouldn't that be $\displaystyle { n + 2 \choose 3}$ ?

A Former Brilliant Member - 5 years, 6 months ago

Right! Sorry about that. Thanks for catching my mistake. :)

Brian Charlesworth - 5 years, 6 months ago

Nice problem and solution

Dev Sharma - 5 years, 6 months ago

I don't understand how you turned $1+...+n$ into a binomial coefficient $\binom{n+1}{2}$ . I understand how you used the hockey-stick identity to find the sum, but I don't understand what made you think pascal's triangle.

Jason Simmons - 5 years, 6 months ago

If you know the hockey-stick identity, consider this:

The numbers 1, 2, 3, ..., $n$ form the left diagonal $\left(\begin{array}{c} n \\ 1 \end{array}\right)$ . You can use the hockey-stick theorem for that, too, and get $\left(\begin{array}{c} n+1 \\ 2 \end{array}\right)$ ...!

Arjen Vreugdenhil - 5 years, 6 months ago

Ah! I see it now. Nice problem by the way!

Jason Simmons - 5 years, 6 months ago

Jason Simmons
Dec 12, 2015

Summing the amount of gifts for each day for the first five days we see the sequence $a_n=1, \: 3, \: 6, \: 10, \: 15, ...$ That is to say that we obtain one gift on the first day, three gifts on the second day, six gifts on the third day, ten gifts on the fourth day, fifteen gifts on the fifth day, and so on. We can readily see that this is a quadratic sequence such that $a_n=1, \: 3, \: 6, \: 10, \: 15, ..., \; \frac{n^2}{2}+\frac{n}{2}$ From here we can simply find the sum of this sequence from the first day of Christmas to the twelfth day of Christmas. This leaves us with the sum

$\frac{1}{2} \sum_{n=1}^{12} n^2+n=\boxed{364}$

Pulkit Gupta
Dec 11, 2015

The given problem can be represented by this summation

(1) + ( 1+2) + ( 1+2+3)+......+( 1+2+3+4+5+6+7+8+9+10+11+12) = 12 * 1 + 11 * 2 + 10 * 3+...+1 * 12 = $\sum_{i=1}^{12}$ n(13-n) = 364

And a Partridge, une Perdrix

The answer is 364.

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