Just. Another Summation

The summation itself is

$\sum_{n=4}^{20} \frac{(n-4)!}{n!}$

$= \sum_{n=4}^{20} \frac{(n-4)!}{n \times (n-1) \times (n-2) \times (n-3) \times (n-4)!}$

$= \sum_{n=4}^{20} \frac{1}{n \times (n-1) \times (n-2) \times (n-3)}$

Notice that

$\frac{1}{(n-1) \times (n-2) \times (n-3)} - \frac{1}{n \times (n-1) \times (n-2)}$

$= \frac{(n) - (n-3)}{n \times (n-1) \times (n-2) \times (n-3)}$

$= \frac{3}{n \times (n-1) \times (n-2) \times (n-3)}$

So

$\frac{1}{n \times (n-1) \times (n-2) \times (n-3)} = \frac{1}{3} \times (\frac{1}{(n-1) \times (n-2) \times (n-3)} - \frac{1}{n \times (n-1) \times (n-2)})$

Then substitute that to the summation

$\sum_{n=4}^{20} \frac{1}{n \times (n-1) \times (n-2) \times (n-3)} = \sum_{n=4}^{20} \frac{1}{3} \times (\frac{1}{(n-1) \times (n-2) \times (n-3)} - \frac{1}{n \times (n-1) \times (n-2)})$

$= \frac{1}{3} \times (\sum_{n=4}^{20} \frac{1}{(n-1) \times (n-2) \times (n-3)} - \frac{1}{n \times (n-1) \times (n-2)})$

$= \frac{1}{3} \times ((\frac{1}{3 \times 2 \times 1} - \frac{1}{4 \times 3 \times 2}) + (\frac{1}{4 \times 3 \times 2} - \frac{1}{5 \times 4 \times 3}) + ... + (\frac{1}{19 \times 18 \times 17} - \frac{1}{20 \times 19 \times 18})$

Use the telescopics, then it will remain

$\frac{1}{3} \times (\frac{1}{3 \times 2 \times 1} - \frac{1}{20 \times 19 \times 18})$

$= \frac{1}{3} \times (\frac{1}{6} - \frac{1}{6840})$

$= \frac{1}{3} \times \frac{1140-1}{6840}$

$= \frac{1}{3} \times \frac{1139}{6840}$

$= \frac{1139}{20520}$

1139 and 20520 are coprime, so $a = 1139$ and $b=20520$ , then $a + b = \boxed{21659}$

*cmiiw

$\begin{aligned} \sum_{n=4}^{20} \frac{(n-4)!}{n!} & = \sum_{n=4}^{20} \frac{1}{\color{#3D99F6}{n}\color{#D61F06}{(n-1)(n-2} \color{#3D99F6}{(n-3)}} \\ & = \sum_{n=4}^{20} \frac{1}{\color{#3D99F6}{n(n-3)}\color{#D61F06}{(n-1)(n-2)}} \\ & = \sum_{n=4}^{20} \frac{1}{\color{#3D99F6}{(n^2-3n)}\color{#D61F06}{(n^2-3n+2)}} \quad \small \color{#3D99F6}{\text{Note: }} \frac{1}{\color{#3D99F6}{n^2-3n}} - \frac{1}{\color{#D61F06}{n^2-3n+2}} = \frac{2}{\color{#3D99F6}{(n^2-3n)}\color{#D61F06}{(n^2-3n+2)}} \\ & = \frac{1}{2} \sum_{n=4}^{20} \left(\color{#3D99F6}{\frac{1}{n(n-3)}} - \color{#D61F06}{\frac{1}{(n-1)(n-2)}} \right) \\ & = \frac{1}{2} \sum_{n=4}^{20} \left[\color{#3D99F6}{\frac{1}{3} \left(\frac{1}{n-3} - \frac{1}{n}\right)} - \color{#D61F06}{\left(\frac{1}{n-2}-\frac{1}{n-1} \right)} \right] \\ & = \frac{1}{2} \left[ \color{#3D99F6}{\frac{1}{3} \left( \sum_{n=1}^{17} \frac{1}{n} - \sum_{n=4}^{20} \frac{1}{n}\right)} - \color{#D61F06}{\left(\sum_{n=2}^{18} \frac{1}{n}- \sum_{n=3}^{19} \frac{1}{n} \right)} \right] \\ & = \frac{1}{2} \left[ \color{#3D99F6}{\frac{1}{3} \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{18}-\frac{1}{19} - \frac{1}{20} \right)} - \color{#D61F06}{\left(\frac{1}{2}-\frac{1}{19}\right)} \right] \\ & = \frac{1}{2} \left[ \color{#3D99F6}{\frac{1}{3} \left(\frac {3420+ 1710 +1140-190-180-171}{3420} \right)} - \color{#D61F06}{\left(\frac{19-2}{38} \right)} \right] \\ & = \frac{1}{2} \left[ \color{#3D99F6}{\frac{1}{3} \left(\frac {5729}{3420} \right)} - \color{#D61F06}{\frac{17}{38}} \right] \\ & = \color{#3D99F6}{\frac {5729}{20520}} - \color{#D61F06}{\frac{17}{288}} \\ & = \frac{1139}{20520} \end{aligned}$

$\Rightarrow a + b = 1139 + 20520 = \boxed{21659}$