What is the probability that two interns were born on the same day (not necessarily in the same year)? Note: There are 62 interns.
We will determine the probability of the opposite, i.e., the probability that we cannot find two people born on the same day. In this case, we need to choose 62 different days from the 365 days of the calendar, with order since two interns are not interchangeable: this is an arrangement. Therefore, there are $\frac{365!}{(365-62)!}$ cases in which we cannot find two interns born on the same day. The total number of possibilities is $365^{62}$. Thus, the probability that two interns were born on the same day is $1-\frac{1}{365^{62}} \times \frac{365!}{(365-62)!}=99.59095749\%$.

Proposition 1.12. There are $\frac{n!}{k!(n-k)!}$ ways to choose $k$ elements from $n$ without considering the order. We denote this quantity as $\binom{n}{k}$.

Proof. We know that there are $k!$ ways to order the $k$ chosen elements. Therefore, when counting the number of ways to choose $k$ elements from $n$ with order, we count $k!$ times too many. Thus, we need to divide $\frac{n!}{(n-k)!}$ by $k!$, and we get $\frac{n!}{k!(n-k)!}=\binom{n}{k}$.

Proposition 1.13 (Binomial Theorem).

$$
(a+b)^{n}=\binom{n}{0} a^{n} b^{0}+\binom{n}{1} a^{n-1} b^{1}+\binom{n}{2} a^{n-2} b^{2}+\ldots+\binom{n}{k} a^{n-k} b^{k}+\ldots+\binom{n}{n} a^{0} b^{n}
$$