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.

Since $7$ is a factor of $\frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
