Suppose, there are 3 "e" . I want to select 2 e from these 3 e. By common sense, we can say that we can select by only one way...But I want to calculate using combination formula...
MY APPROACH:
During the selection, 2 e will always occur. So, we can calculate excluding 2 e... n∧C_r-2 Where, n=3 and r=2. So, the answer is 1.
Is my approach correct?
If you want to number all the ways you can choose $k$ elements from a group of $n$ you have to use the binomial coefficient.
In your case the result is: $$\binom{3}{2}=\binom{3}{3-2}=\binom{3}{1}=3$$
But since all the three elements are identical, you have just $1$ way to choose 2 of them. In fact they are undistinguishable.