Solving Equations with Factorials I am attempting to solve an equation ${{n-2} \choose {2}} + {{n-3} \choose {2}} + {{n-4} \choose {2}} = 136$. With the formula for a combination being $\frac{n!}{r!(n - r)!}$, I simplified the given equation to:
$(n-2)! + (n-3)! + (n-4)! = 272n - 1088$ 
However, I am not sure how I would solve for $\\n$ in the simplified version. I have looked into Sterling's Approximation, hoping to find some way to solve this there, but I believe that is going about it incorrectly.
Any help that can be offered in solving this equation is greatly appreciated.
 A: Note that
$$\dbinom{n-2}2 = \dfrac{(n-2)(n-3)}{2}$$
$$\dbinom{n-3}2 = \dfrac{(n-3)(n-4)}{2}$$
$$\dbinom{n-4}2 = \dfrac{(n-4)(n-5)}{2}$$
Hence,
$$\dbinom{n-2}2 + \dbinom{n-3}2 + \dbinom{n-4}2 = \dfrac{(n-2)(n-3)}{2} + \dfrac{(n-3)(n-4)}{2} + \dfrac{(n-4)(n-5)}{2}$$
Now solve the quadratic to get the answer.
EDIT
$$\dbinom{m}2 = \dfrac{m!}{2! (m-2)!} = \dfrac{m(m-1) \cdot (m-2)!}{2! \cdot (m-2)!} = \dfrac{m(m-1)}2$$
A: It is best to leave factorials out of this: $\dbinom{x}{2}=\dfrac{x(x-1)}{2}$.
Remarks: $1.$ The binomial coefficients were simplified incorrectly. 
$2.$ We don't even need "algebra" to solve the problem. Evaluate your expression, for $n=4, 5, \dots$. After a short while your sum is clearly too big. 
$3.$ We describe a way to calculate a binomial coefficient, say $\dbinom{10}{3}$. By the formula you used, this is $\dfrac{10!}{3!7!}$.
Look at the top. This is $(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)$. Note that it is equal to $(10)(9)(8)(7!)$. Now the $7!$ part cancels the $7!$ at the bottom. 
So our binomial coefficient is equal to $\dfrac{(10)(9)(8)}{3!}$. We did not have to calculate a big factorial.
For something like $\dbinom{x}{2}$, use $\dfrac{x!}{2!(x-2)!}$. The top is $x(x-1)(x-2)!$. The $(x-2)!$ cancels the one at the bottom, and we end up with $\dfrac{x(x-1)}{2!}$.
A: Bad simplification, that one.
$$
\begin{align*}
\binom{n - 2}{2} + \binom{n - 3}{2} + \binom{n - 4}{2}
  &= \frac{(n - 2)(n -3)}{2} + \frac{(n - 3) (n - 4)}{2} + \frac{(n - 4)(n - 5)}{2} \\
  &= \frac{3 n^2 - 21 n + 38}{2}
\end{align*}
$$
The binomial coefficient $\binom{n}{k} = \binom{n}{n - k} = \frac{n (n - 1) \ldots (n - k + 1)}{k!}$ ($k$ factors upstairs and downstairs) in ubicuous in combinatorics.
You surely can take it from there...
A: $n=13$  after solving this equation $3n^2-21n-234=0 
$
