How to isolate variable algebraically in a combinatorics equation? How would I isolate a variable algebraically in a combinatorics equation? 
For example, if I'm given:
$$C(k, 2) = 45$$
How would I solve for $k$, without trying random values of $k$? I know that $$C(k, r) = \frac{k!}{r!(k-r)!}$$ but I'm unsure how I'd solve for $$ \frac{k!}{2!(k-2)!} = 45$$ Or if this is even the right approach. A smaller question that might solve the larger would be, how do I get rid of factorials algebraically? 

The number $n!$ can be defined recursively as follows:


*

*$1! = 1$

*$n! = n(n - 1)!$, $n \geq 1$


Using the definition to compute $5!$ yields
\begin{align*}
5! & = 5 \cdot 4!\\
   & = 5 \cdot 4 \cdot 3!\\
   & = 5 \cdot 4 \cdot 3 \cdot 2!\\
   & = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1!\\
   & = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\end{align*}
Notice that if $n \geq 2$, we have $n! = n(n - 1)! = n(n - 1)(n - 2)!$.  Hence,
$$\binom{n}{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n(n - 1)}{2}$$
Thus,
\begin{align*}
\binom{n}{2} & = 45\\
\frac{n!}{2!(n - 2)!} & = 45\\
\frac{n(n - 1)(n - 2!}{2(n - 2)!} & = 45\\
\frac{n(n - 1)}{2} & = 45\\
n(n - 1) & = 90\\
n^2 - n & = 90\\
n^2 - n - 90 & = 0\\
(n - 10)(n + 9) & = 0
\end{align*}
The quadratic equation has solutions $n = 10$ and $n = -9$.  However, it is not possible to select two objects from $-9$ objects, so $n = -9$ is an extraneous solution.  Hence, $n = 10$.  
If $n \geq 3$, we have $n! = n(n - 1)(n - 2)(n - 3)!$.  Thus,
$$\binom{n}{3} = \frac{n!}{3!(n - 3)!} = \frac{n(n - 1)(n - 2)(n - 3)!}{3!(n - 3)!} = \frac{n(n - 1)(n - 2)}{3!}$$
so you could solve the equation
$$\binom{n}{3} = 56$$
by solving the cubic equation
$$\frac{n(n - 1)(n - 2)}{3!} = 20$$
More generally, 
\begin{align*}
\binom{n}{k} & = \frac{n!}{k!(n - k)!}\\ 
             & = \frac{n(n - 1)(n - 2) \cdots (n - k + 1)(n - k)!}{k!(n - k)!}\\
             & = \frac{n(n - 1)(n - 2) \cdots (n - k + 1)}{k!}
\end{align*}
