If there are 100 people in a room then what is the probability that at least 2 of them share birthdays? I read that answer should be $1-\left(\frac{365!}{265!\times365^{100}} \right)$, and I understood why it is so.
However if I solve it another way I am getting the wrong answer. If we think of the problem as assigning birthdays to people, then there are ${365 \choose 100}$ ways to assign each person with a different birthday and ${464 \choose 100}$ ways so that one birthday could be assigned to more than one person. So why the following equation gives me the wrong answer: $$1-\frac{{365 \choose 100}}{{464 \choose 100}}$$
I don't understand why the order of people matters in this question.
The problem is that the persons are distinguishable objects.
To understand the flaw in your reasoning consider this extreme example. According to your solution there's only one configuration in which all people are born on 1 January, which is right. However again according to your solution there is a single configuration in which one person is born on 2 January and the rest are on 1 January. But in reality there are 100 such configurations, as the the first person can be born on 2 January, then the second and so on.
To illustrate it even better according to your solution the probability that all people are born on 1 January is same as the probability that they are born on 1 January, 2 January, 3 January ... Now imagine the first person coming then there's a $\frac{1}{365}$ chance the first configuration is satisfied, but there's $\frac{100}{365}$ the second one is satisfied. Similarly for the second person you have $\frac{1}{365}$ chance the first confifuration is true, while $\frac{99}{365}$ the second one is.