# Solution

Prove that in any group of $n \ge 2$ people, there are at least two who are acquainted with the same number of people.

Let's argue indirectly...

Assume that all are acquaintances with a different number of people.

Notice, there are $n$ people and $n$ possibilities for the numbers of acquaintances each can have: $0,1,2,\ldots,n-1$.

If each person is acquainted with a different number of people, then each of these numbers must get used.

As such, there must be a person acquainted with everyone else ($n-1$ people), as well as a person with no acquaintances (0 people).

However, the person acquainted with everyone else can't be acquainted with the person who has no acquaintances, so unless $n-1=0$, which makes these two values associated with the same person (which happens when $n=1$) we have our contradiction.

As such, when $n \ge 2$, we must reject our assumption.

Instead, its opposite must be true: There must be two people with the same number of acquaintances.

QED.