Given 101 Colored Balls, prove that there are 11 balls of the same color or 11 balls of different colors.
My approach: I'm assuming that there are no 11 balls of the same color then trying to prove that there are 11 balls of different color and vice versa.
I feel like I have intuition to the solution, like in the first direction (what I stated above) If I try to divide the balls of the same color into groups I need at least 11 groups
I'm struggling to write stuff formally, any suggestions on how to apply the pigeonhole principle in this question?
Group each type of colored ball in groups of $10$
$\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{10}\boxed{?}$
Then the $101th$ colored ball has to be either the same as one of the $10$ colors placed or else a $11th$ color in the box marked $?$
Thus given $101$ Colored Balls,it is proved that there are either $11$ balls of the same color or $11$ balls of different colors.
Response to OP's comment
If we go strictly by the original definition of the pigeonhole principle, each hole can accomodate only one pigeon, but we can extend it, as I did in the answer for simplification, putting $10$ balls of one color in one hole.
If you want more formality, have $100$ holes and fill the holes to the maximum allowed of each color to reach $100$, and thus prove that the $101th$ pigeon is necessarily either one from already placed $10$ colors, or a new colored pigeon.
I don't see any need to make it more "formal"