Factorial Fire

Let $\displaystyle S = \dfrac 1{2!17!} + \dfrac 1{3!16!} + \dfrac 1{4!15!} + \cdots + \dfrac 1{9!10!} = \sum_{k=2}^9 \frac 1{k!(19-k)!}$ . Then let us consider:

$\begin{aligned} (1+x)^{19} & = \small 1 + 19x + \frac {19!}{2!17!}x^2 + \frac {19!}{3!16!}x^3 + \cdots + \frac {19!}{9!10!}x^9 + \frac {19!}{10!9!}x^{10} + \cdots + \frac {19!}{16!3!}x^{16} + \frac {19!}{17!2!}x^{17} + \cdots + 19x^{18} + x^{19} \\ 2^{19} & = \small 1 + 19 + \frac {19!}{2!17!} + \frac {19!}{3!16!} + \cdots + \frac {19!}{9!10!} + \frac {19!}{10!9!} + \cdots + \frac {19!}{16!3!} + \frac {19!}{17!2!} + \cdots + 19 + 1 \quad \quad \color{#3D99F6} \text{Putting }x=1 \\ 2^{19} & = 40 + 2(19!) \sum_{k=2}^9 \frac 1{k!(19-k)!} \\ 2^{19} & = 40 + 2(19!)S \\ \implies S & = \frac {2^{19}-40}{2(19!)} = \frac {2^{18}-20}{19!} \end{aligned}$

$\implies a+b+c = 18+20+19 = \boxed{57}$

Relevant Wiki: Pascal's Triangle

We have to find $S=\displaystyle\sum_{n=2}^9 \dfrac{1}{n!(19-n)!}$

Let's first consider

$\displaystyle\sum_{n=0}^{19} \dfrac{19!}{n!(19-n)!}=2^{19}$

Since Pascal's triangle is symmetrical

$\displaystyle\sum_{n=0}^{9} \dfrac{19!}{n!(19-n)!}=2^{18}$

Manipulating this formula

$\begin{aligned} & \sum_{n=0}^{9} \dfrac{1}{n!(19-n)!}=\dfrac{2^{18}}{19!}\\ \Longrightarrow & \dfrac{1}{19!}+\dfrac{1}{18!} +\mathbin{\color{#D61F06} \sum_{n=2}^{9} \dfrac{1}{n!(19-n)!}}=\dfrac{2^{18}}{19!}\\ \Longrightarrow & \mathbin{\color{#D61F06} S}=\dfrac{2^{18}}{19!}-\dfrac{1}{19!}-\dfrac{19}{19!}\\ \Longrightarrow & S = \dfrac{2^{18}-20}{19!}=\dfrac{2^a-b}{c!}\\ \Longrightarrow & a+b+c=18+20+19=\boxed{57} \end{aligned}$