Of 100 people seated at a round table, more than half are women. Prove that there exist two women who are seated diametrically opposite each other.

Question

Of 100 people seated at a round table, more than half are women. Prove that there exist two women who are seated diametrically opposite each other.

Answer 1

There are $50$ pairs of diametrically opposite seats. You have more than $50$ women.... Does this make you think of stuffing birds in holes?

Answer 2

At a round table with 100 seats, there are 50 pairs of seats diametrically opposite to each other in the pair. The number of women $n$ to distribute among the pairs is greater than the number of pairs, $m$; i.e., $n > m$. By the pigeonhole principle, at least one pair contains more than one woman.

Answer 3

First of all, let us consider that no two such women exist, as mentioned in the question.

Say, we consider any $2$ diametrically opposite sitting people as an ordered pair. So we will have $50$ such ordered pairs as for example, $(P_1,P_{51}),(P_2,P_{52}),(P_3,P_{53}),(P_4,P_{54}), \ldots (P_{50},P_{100})$.

If every one persons from those $50$ pairs have at least $1$ man, then you will have at least $50$ man among the $100$ people.

So no. of woman can be at most $50$.

Hence no. of women $\not >$ no. of men.

Contradiction.