How would I solve $\frac{(n - 10)(n - 9)(n - 8)\times\ldots\times(n - 2)(n - 1)n}{11!} = 12376$ for some $n$ without brute forcing it? Given this equation:
$$
\frac{(n - 10)(n - 9)(n - 8)\times\ldots\times(n - 2)(n - 1)n}{11!} = 12376
$$
How would I find $n$?
I already know the answer to this, all thanks toWolfram|Alpha, but just knowing the answer isn't good enough for me. I want to know on how I would go about figuring out the answer without having to multiply each term, then using algebra to figure out the answer.
I was hoping that there might be a more clever way of doing this.

Using @pharmine's factorization, we have
$$\begin{eqnarray*}
n(n-1)(n-2)\cdots(n-10) &=& 17\cdot 13 \cdot 7 \cdot 2^3 \cdot 11! \\
&=& 17\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2
\end{eqnarray*}$$
Now, $17$ divides the LHS, so LHS consists of eleven consecutive integers including a multiple of $17$; given the size of the product, that multiple is likely "$17$" itself. The prime $19$ isn't a factor, so the consecutive integers are either "$18$" through "$8$" (that is, $n=18$), or "$17$" through "$7$" ($n=17$). In either case, the list of integers contains "$17$" through "$8$"; constructing these uses up the RHS factors:
$$(17)\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2$$
$$(17\cdot 16)\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 7\cdot 6\cdot 5\cdot 4\cdot 3$$
$$(17\cdot 16 \cdot 15)\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 7\cdot 6\cdot 4$$
$$\cdots$$
$$(17\cdot 16 \cdot 15 \cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8) \cdot 7$$
... and we conclude that $n=17$.
