One way to look at it this:

Had the arrangement been \(E_1E_2L_1L_2\) (with both Es and Ls different), the number of arrangments would have been 4P2 = 4!/(4-2)!=12, with nPr=n!/(n-r)!

But the 2 Es and the 2 Ls are the same. Thus, the number of arrangements will be:

Arrangements with 2 Es + Arrangements with 2 Ls + Arrangements with 1 E and 1 L

2P2/2! + 2P2/2! + 2P2 = 1+1+2 = 4

The first 2 expressions are divided by 2! as both the Es are same. If we had both the Es different (\(E_1\) and \(E_2\)), then we would not have divided the expressions by 2!

The concept remains the same.

An alternate way to take a look at the above problem (2 out of 4) is to pick 1 combination and see in how many ways you can permute the other places.

Lets start with 1 possible combination: EL

E_ with the '=' representing another letter. We have the choice of E,L (I can only count 2 Ls as 1 L as both of them are indistinguishable).

So number of ways to choose 'E' out of 2 Es: 1

Number of ways to choose the remaining letter: 2 (remaining E and 1 L, as both of them are same!)

Thus E_ can provide us 2 arrangements. Now, we can choose either of E or L to be the first letter and thus we need to multiply 2 arrangements obtained from E_ to complete all the possible arrangements for a total of 2*2 = 4 arrangements.

This one is straightforward as the number of repetitions is less than the total length of the string (3 in this case) required. Thus the required number of arrangements are:

4P3/(2!*2!) = 6 possible arrangements.

EEL
ELE
LEE
LLE
LEL
ELL

Yes, your method of counting different possible arrangements seems to make sense to me. Sometimes, mere application of formulae isn't straightforward unless we understand the different arrangements possible/allowed.

The first method that you proposed isn't correct , IMO. When you write 6p4/ (3!x3!) you are not counting all the possible arrangements.

I think Chetan2u has explained nicely above. For selecting 2 out of 4 with 2 reps each of E and L needs to be treated differently from 3 out of the same 4 as you will need to take 1 from the second letter to make combination of 3 letters.

The same scenario applies for selecting 4 out of 6 with 3 each of E and L.