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.

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.
