Solving for $r$ in ${12\choose{r}}=924$ I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically?  I can't seem to simplify the combination expression without using a large number of cases.
 A: Since $7$ is a factor of $\frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
A: I don't know of a quick way.  As $12-r$ is a solution whenever $r$ is, you can concentrate on $r \le \frac {12}2$  For reasonably small values, you can then do binary search.
If the value $12$ is larger, you can use Stirling's approximation  $12Cr=\frac {12!}{r!(12-r)!}\approx\frac {12^{12}\sqrt{24\pi}}{r^r(12-r)^{12-r}2\pi\sqrt{r(12-r)}}$  Now take the logs and solve the equation to get $r=$ and expression involving $\log r, \log(12-r)$. Since the log changes rather slowly, guess that $r \approx \frac {12}4$ (the same starting point as the binary search.  Plug that value in on the right, compute $r$, and you should come close with few iterations.  Once you get close, use the fact that $r$ is a natural.
A: Use the definition $$_nC_r = \frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$\frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2\cdot 3 \cdot 7 \cdot 11$ which means $$\frac{12!}{924} =\require{cancel} \frac{1\cdot 2\cdot \cancel{3} \cdot \cancel{4} \cdot 5 \cdot 6 \cdot \cancel{7} \cdot 8 \cdot 9 \cdot 10 \cdot \cancel{11} \cdot 12}{\cancel{2^2}\cdot \cancel{3} \cdot \cancel{7} \cdot \cancel{11}}  \\ = 2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 10 \cdot 12$$ So now you have $$2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 10 \cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
A: You can spot the high prime factors of $924$ - which exist by Bertrand's Postulate. 
Here you see $11$ and $7$ and the highest value for $r$ which avoids cancelling $7$ is $6$, which happens to be the only possibility.
This would work more generally to reduce the possbilities. Note that if a high prime doesn't appear in the factorisation that puts a constraint on too, because that prime has to cancel.

This puts the larger of $r$ and $n-r$ greater than or equal to the highest prime less than $n$ excluded from the factorisation and lower than any prime greater than $\frac n2$ which is included.
