<s>[INST] 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.
 [/INST] 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
$$\dbinom{m}2 = \dfrac{m!}{2! (m-2)!} = \dfrac{m(m-1) \cdot (m-2)!}{2! \cdot (m-2)!} = \dfrac{m(m-1)}2$$
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