Imagine n slots with n-1 spaces between them. Putting k-1 separators into the spaces between the slots (max 1 separator per slot) determines a valid selection: put first color into all of the slots before the first separator, the second color into the slots after the first and before the second separator, etc. Since there's at least one space between each two separators, there will be at least one pencil in each color.
The mapping is one-to-one, since every selection also generates a unique configuration of separators.
Putting k-1 separators into n-1 spaces can be done in N(n-1,k-1) ways, where N is the Newton symbol, so the answer is N(n-1,k-1).
Another way of representing this, based on djna answer:
Fix k pencils, one in each color - you nead at least one in each color, so these will make sure this requirement is fulfilled. Now you have n-k picks left, which you can split between colors anyway you want, this time not having to care about picking each color (this is assured by k pencils chosen first.) The number of solutions is the number of ways you can split n-k indistinguishable pencils into k distinguishable(*) partitions.
How to enumerate this? Add k-1 pens to your n-k pencils, line them up and color from left to right, changing color after encountering a pencil. For example, representing pens as * and pencils as -, with three colors (red, green, blue) this:
--**-
represents two red pencils, zero green and one blue. In such a representation, there are (n-k) + (k-1) = n-1 elements (pens and pencils.) From these, you need to pick k-1 pen positions (or n-k pencil positions, since picking one set determines the other.) The number of ways you can do that is N(n-1,k-1).
(*) I assume that "2 red, one green" is different from "2 green, one red", otherwise is a totally different task.