Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.

Question

There are two ways to prove it. One way is...

Consider $41$ chess pieces on $10$ board rows. By using pigeonhole princ., there must be one row that has at least $\lceil{\frac{41}{10}}\rceil$ $=$ $5$ (but less than $10$) pieces. Eliminate that row. Then there are $9$ rows and at least $31$ chess pieces remaining. By using pigeonhole princ., there must be one row that has at least $\lceil{\frac{31}{9}}\rceil$ $=$ $4$ pieces..........

As you can see, this proving way is really long, you have to eliminate rows by rows and choose one piece in one of the eliminated rows to prove and conclude.

Can you find the other way to prove by using the pigeonhole principle for only one time?

Answer 1

$4$ rows can at most provide room for $4 \cdot 10 = 40$ pieces, So, with $41$ pieces, you'll need need to use at least $5$ rows to fit all the pieces, meaning that there have to be $5$ pieces that are on $5$ different rows (I assume that that is what the question is asking you to prove)

Answer 2

Since we have $10$ colulmns we must have, by Pigeonhole principle, in one of them at least $5$ chess pices. So they are not in the same row and we are done