Mathematics: Combinations - Creating a committee?

[Nikki T]

A class consists of 12 boys, 5 whom are prefects. How many committees of 8 can be formed if the committee is to have:
a)3 prefects?
b)and least 3 prefects?

I understand part a, where you take the combination of prefects and multiply it by the combination of the rest of the students.[(5C3 = 10 different ways to have a different combination of three of the five prefects) times (7C5 = 21 different ways to have a different combination of the remaining non-prefect boys) to result in the 210 combinations the book says I should end up with]

Part b is the problem. When it says at least three prefects, that means the rest of the boys (including the other two prefects, who had to be left out in part a) can have an equal chance at the 5 remaining spots in the committee. When I find the combination of the nine leftover boys and multiply it by the ways of having three prefects, it does not result in 420, the answer given, but rather 1260.

How do I solve part b?