The pigeonhole principle - 30 pens in a drawer I need help with the following task. It needs to be solved using the pigeonhole principle. 
There are 10 red, 8 blue, 8 purple and 4 yellow pens in a drawer. We pick them out, one by one, in the dark. What is the least number of pens that we need to pull out if we want to ensure that we have 
a) at least 1 pen of each color
b) at least 6 blue pens?
I realize for example that if we want at lest 4 pens of each color, we need to pull out at least 11 ones, but I do not understand how to get one of each, or a specific number of a given color. I would solve this using statistics, but it has to be the principle.

Think logically.
Since you are drawing pens in the dark, it is very much possible that you require the maximum possible number of draws, i.e.  
For $1^{st}$ case:
You might start off with Red, then finish off all Reds, Blues and Purples, i.e. $10+8+8=26$ draws in the worst case. Now, only left with Yellow, just draw one.
So, at most you may need $27$ draws.
While, if lucky enough, you may grab $4$ different colors in just $4$ draws.
So, to be sure, you will need at least $27$ draws.  
For $2^{nd}$ case:
Finish drawing all other colors, contributing to $10+8+4=22$ draws. Now, left with only Blues, just draw $6$, so at most $22+6=28$ draws in the worst-case.
Again, in the rarest case, you may need just $6$ draws; though its probability is quite less.
So, to be sure, you will need at least $28$ draws.   
Try to convince yourself!
