May I know why is the fact that P(The best is among the first n) is calculated in this way?

Question

I am trying to solve this problem:
On the basis of an interview, the N candidates for admission to a college are ranked in order according to their mathematical potential. The candidates are interviewed in random order(that is , each possible order is equally).
Find the probability that the best amongst the first n candidates interviewed is the est or second best overall.
So，what I do here is that we know that:
P(the best is among the first n)=$\frac{n}{N}$ So, I think that the the total probabilities can be calculated in this way:
$$2\times \frac{n}{N}(1-\frac{n}{N}) + \frac{n}{N}$$ The reason I am doing this is that I we need to add the probability of the second best is in the first n but the best overall is not and the best is in the first n. So , we can get the answer.
After that I went on to check the mark scheme.
It said that the we should calculate P(the second is in the first n) + P(the best is in the first n) - P(both of them are in the best n)
I understand why the mark scheme is doing this, but the way they are doing this I couldn't understand. what they did are:
$$2\times \frac{(N-1)C(n-1)}{NCn} - \frac{(N-2)C(n-2)}{Ncn}$$ I couldn't understand the expression for it, could you guys pls explain it to me. thank you very much

Answer 1

Well, I believe that the formula arises because you know who are in the top two, so, to count in how many configurations each must appear, is the same as to count all possible $n$-subsets from $N$ people knowing who's going to be one of them: $\binom{N - 1}{n - 1}$. The second term follows a similar approach.