Twelve Days of Christmas

Let the total number of gifts the true love will send in $n$ days be $S(n)$ , then we have:

$S(n) = \sum_{i=1}^{n}{\sum_{i=1}^{n}{i}}$

$\quad = \sum_{i=1}^{n}{\cfrac{i(i+1)}{2}}$

$\quad = \frac{1}{2} \sum_{i=1}^{n}{(i^2+i)}$

$\quad = \frac{1}{2} \left( \sum_{i=1}^{n}{i^2} + \sum_{i=1}^{n}{i} \right)$

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

$\quad = \frac{1}{12} \left( n(n+1)(2n+1) + 3n(n+1) \right)$

$\quad = \frac{1}{12} n(n+1)(2n+4)$

$\quad = \frac{1}{6} n(n+1)(n+2)$

For $n=12$ , then $S(12)=\frac{1}{6}(12)(13)(14)=\boxed{364}$ .

a lenghty one...

Got it right,but how is this a question of combinatorics ? seems to be of progressions.

On first day the someone's true love sends one gift. On second day he sends 2 + 1 = 3 gifts and on third day he sends 3 + 2 + 1 = 5 gifts and so on. On $n^{th}$ day he sends $\frac{n(n+1)}{2}$ gifts. Total number of gifts sent is $\sum_{n = 1}^{12} \frac{n(n+1)}{2} = 364$