How many 2 letter arrangements possible with "EELL"?

Question

Hi All,

I was learning Combinatorics in MGMAT guides, when I came up with this problem myself: How many 2 letter arrangements are possible with "EELL"?

I know there are 4 arrangements as I have shown below. But using Combinatorics properties, I don't know how to solve.

EE, EL, LE and LL.

Thanks,
Rudra

ENGRTOMBA2018 · Answer

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!

rudra07 · Answer

Hi, thanks for the reply. It helped but I still have some questions in mind. You can help me if you kindly solve for 3 letter arrangements for letters "EELL". So the rephrased question would be: how many 3 letter arrangements are possible with "EELL"?

ENGRTOMBA2018 · Answer

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

rudra07 · Answer

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?

"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?

ENGRTOMBA2018 · Answer

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.

chetan2u · Answer

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

ENGRTOMBA2018 · Answer

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.

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

rudra07 · Answer

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

How many 2 letter arrangements possible with "EELL"?

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