How many letter combinations can be composed of letters of

How do you differentiate between the same letter L? I thought about the number of ways of picking one L and one E out of 2 L's and 2 E's respectively.

Write down and see if you really get 24. In the end L1=L2 & E1=E2 and your combinations will reduce down to 6

Ah! you are right icandy - thanks.
kudos given

Can somebody elaborate on this problem,
I can not get which method you guys are using to come to the solution.

Slot method or using specific combinatorics formula?

My thinking was that Letter L could be chosen in two ways, Letter E in 2 ways as well, and the middle V only one way, thus
I am getting 4
I know it is not correct answer.
Thank you.

How many letter combinations can be composed of letters of the word LEVEL if all these combinations have to begin with L and end with E?

Keep it simple:

\(L---E\): three distinct letters are left - V, E and L. We can place them in three slots in 3!=6 # of ways (basically it's the same as # of permutations of 3 distinct letters - 3!).

Thank you,
Now I got it!

Check he differnce between the two questions

How many letter combinations can be composed of letters of the word LEVEL if all these combinations have to begin with L and end with E?

* 4
* 6
* 10
* 12
* 24

How many ways can the letter - TRUST - be arranged?

* 30
* 60
* 6
* 3
* 24

Ans to both is B

Hi Bunuel,
Isn't the question ambiguous? At its current form, why should we not include word combinations which would have just one or two letters in between L and E. For example, should not LVE or LLVE be included? Dont see anything in the question which would stop us from including such combinations. Am I wrong? Please help. Thanks.

As I read the question, you're perfectly right - there's no reason from the wording to assume we need to use all 5 of the letters (though if you don't need to use all five letters, the right answer is not among the choices).

Perhaps worse still, the question asks how many 'combinations' we can make. In mathematics, a 'combination' is a selection *where order does not matter*. If we are creating 'words', obviously order does matter. It's crucially important that the question be clear about whether order matters here, and it isn't. The language of the question is horrible; the original post is so old I doubt we'll ever know the source, but whatever it is, I wouldn't recommend using it.

Ian ,

dont we have to consider chosing one L out of two and one E out of two for the edges? how come answer is its just 6! then?

Please clarify.

I'm afraid I don't understand your question - what do you mean by 'edges'?

i meant start and end.

we have 2 L's and 2 E's right? Let me know if this approach is right?

selecting one L out 2 * 3 distinct letter permutations * selecting one E out of 2

= 2c1 * 3! * 2c1

finally we divide this by 2! *2! for because we have 2 L's and 2 E's

= 2c1 * 3! * 2c1 / (2! *2!) = 6