Too many factorials!

We can write that as:

$\frac{99!(100+1)98!(99-1)97!(98+1)\cdots(2+1)}{99!(100-1)98!(99+1)97!(98-1)\cdots(2-1)}$

And then we get

$\frac{101\times \cancel{98}\times \cancel{99}\times \cancel{96}\times \cancel{97}\times \cancel{94}\times \cdots \times \cancel{4}\times \cancel{5}\times 2\times \cancel{3}}{\cancel{99}\times 100\times \cancel{97}\times \cancel{98}\times \cancel{95}\times \cancel{96}\times \cdots \times \cancel{6}\times \cancel{3}\times \cancel{4}\times 1} \\ = \frac{101\times 2}{100} \\ = 2\frac{1}{50}$

So, $a = 2$ , $b = 1$ , $c = 50$ and their product is $2\times 1\times 50 = \boxed{100}$