smc

Using complementary probability, we can reduce the problem into two cases- the third shiny penny is drawn on the third draw, or on the fourth draw. For the first case, there is only one way to have the shiny pennies as the first three coins drawn, out of a possible 35 drawings (7 choose 3). For the second case, the third shiny penny has to be the fourth penny drawn, which leaves three possible orderings for the first three coins drawn (NS S S, S NS S, S S NS), out of 35 (7 choose 4). Adding these two probabilities together gives $\frac{4}{35}$, and subtracting this from one yields $\frac{31}{35}$, which makes a 31 and b 35, which sum to $\boxed{66}$.