How many 2 letter arrangements possible with "EELL"?

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

Thanks again. I tried to apply the logic you gave me (the straightforward one) to solve 4 letter arrangements for "EEELLL". I put the formula 6P4/(3!*3!) = 10. But surely I was doing something wrong as I calculated more than 10 arrangements on hand. So I tried for the following 3 possibilities using your first example:

"3 Es and 1 L"-> 4!/3!*1! = 4
"3 Ls and 1 E"-> 4!/3!*1! = 4
"2 Es and 2 L"-> 4!/2!*2! = 2

I got total = 14. I found this worked.

Do you agree with me?

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.

Hi,
why your direct formula for 3 out of 4 worked in EELL was because if you pick up any three out of the 4, it will have 2 of one type and 1 of other type..
But it is not the same case in picking 4 out of 6 in EEELLL..
here it can have two of each type or 3 of one and one of other..
so yoy will have to separate the same types and find answer for each type..
hope it helps

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.

Yes, thanks much guys. That all helped a lot.