There are 10 red, 8 blue, 8 green & 4 yellow pencils inside a box.
How many pencils must be selected at least, so we can be sure that there is one pencil of each colour among them (selected pencils)? Suppose, that we selected pencils in dark.
This exercise is selected to be done by the Generalized pigeonhole principle.
But, I can´t find way to apply rule of the Generalized pigeonhole principle, which says that
for minimum number of objects (N), at least (r) of them, must be in one of (k) boxes, when these objects are distributed among the boxes
in words that least r number of object must be in one of k boxes. I think instead of one, there should be all of boxes.
The number of selected objects (pencils) N, that we have to select among total of 30 pencils is unknown.
r is 1 (because we have to find one pencil of each colour among them).
Number of boxes (k) is 4 (one for each colour).
So, how to find N ?
Thank you.
Don't be caught up in the definition - think of it using basic logic. Come up with a worst-case scenario. We could draw the $10$ red pencils, followed by the $8$ blue pencils, followed by the $8$ green pencils, and without ever drawing a yellow pencil, we have already used $10 + 8 + 8 = 26$ moves. But we still need one more draw to get all the colors. Our answer is thus $26 + 1 = \boxed{27}.$