Proof that there are at least 10 colored cells in a column with the pigeonhole principle

Question

I have a table with 13 rows and 10 columns, and I have to prove that if every row has at least 7 colored cells there will be a column with at least 10 colored cells (with pigeonhole principle).

From what I understand there are 10C7=120 ways to choose cells in each row, but do I relate it to the columns?

Answer 1

Since every row has at least $7$ colored cells, we have at least $7\times 13 = 91$ colored cells in total. So, by the pigeonhole principle, there will be a column in the $10$ columns with at least $\lceil 91/10 \rceil = 10$ colored cells.

Answer 2

Another way of seeing it: the largest allocation such that there are fewer that 10 colored cells/column, is to put 9 in each column 9x10=90, but we have 91, so there is guaranteed to be at last 1 column with 10 colored cells