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? 
 A: Completing the square gives,
$$\frac{k!}{(k-2)!} = k(k-1) = k^2-k+\frac{1}{4} = 2!(45)+\frac{1}{4}.$$
This gives
\begin{align*}
\left(k-\frac{1}{2}\right)^2 &= \frac{361}{4}\\
k-\frac{1}{2}&=\sqrt{\frac{361}{4}}\\
k &= \frac{19}{2}+\frac{1}{2}\\
k&=10
\end{align*}
A: Let us suppose that you know $r$ and want to know $k$ such that $\binom{k}{r}=c$, with $c\in \mathbb{N}$. Let us name $(*)$ this equation.
For $r=0$, $\binom{k}{r}=1$, and you can easly find the solution of $(*)$ if there exists.
If $r=1,2,3,4$ you can solve $(*)$ relatively easy, cause $\binom{r}{2}=\frac{r(r-1)}{2}$, $\binom{r}{3}=\frac{r(r-1)(r-2)}{3!}$ and $\binom{r}{4}=\frac{r(r-1)(r-2)(r-3)}{4!}$, and each one induces an equation of degree 2,3,4, respectively.
But, for $r\ge 5$, the resulting equation doesn't have a general formula cause is of degree $\ge 5$, so you will try with the particular form of $k$ and $r$ in each case. 
Conclusion: For equation $(*)$, only a few cases can be solved in general way. In particular, your original equation becomes $\dfrac{k(k-1)}{2}=45$, so $k=10$
A: 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*}
