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.

$n(n-1)\cdots(n-10)/11! = 2^3 \cdot 7 \cdot 13 \cdot 17$. It's not hard to see that $11! = 2^8 3^3 5^2 7^1 11^1$; this is apparently known as de Polignac's formula although I didn't know the name. Therefore
$n(n-1) \cdots (n-10) = 2^{11} 3^3 5^2 7^2 11^1 13^1 17^1$.
In particular 17 appears in the factorization but 19 does not. So $17 \le n < 19$. By checking the exponent of $7$ we see that $n = 17$ (so we have (17)(16)\cdots (7), which includes 7 and 14) not $n = 18$.
Alternatively, there's an analytic solution. Note that $n(n-1) \cdots (n-10) < (n-5)^{11}$ but that the two sides are fairly close together. This is because $(n-a)(n-(10-a)) < (n-5)^2$. So we have
$$ n(n-1) \cdots (n-10)/11! = 12376 $$
and using the inequality we get
$$ (n-5)^{11}/11! > 12376 $$
where we expect the two sides to be reasonably close. Solving for $n$ gives
$$ n > (12376 \times 11!)^{1/11} + 5 = 16.56.$$
Now start trying values of $n$ that are greater than 16.56; the first one is 17, the answer.
Implicit in here is the approximation 
$$ {n \choose k} \approx {(n-(k-1)/2)^k \over k!} $$
which comes from replacing every factor of the product $n(n-1)\cdots(n-k)$ by the middle factor.
