Total no. of words formed when similar letters are involved Find total number of words formed by arranging the letters in LALLO when:-
1)Two 'L' do not appear together.
2)Three 'L' do not appear together.
My attempt:


*

*First arranging two L such that they do not appear together,so total number of ways equals 1.

*Now arranging two other letters ,so total number of ways-2

*So total words formed =2.

*Same for the second part.


I doubt that my method of solving this is incorrect,plzz tell me how to approach these types of problems.
 A: Your approach for the first problem is great. The nicest problems in combinatorics are when you can describe the thing you are counting with several independent choices. Then you can just multiply the numbers of options for each choice.
For the second one, things are not so clean. If not all three Ls are next to each other, there are still many possibilities for the Ls: L_L_L, or LL_L_, or LL__L, etc. Often when this happens, an easier method is to count the opposite of the condition. Instead, how many words do have all three Ls next to each other? Once you have that number, subtract it from the total number of arrangements of LALLO to get the answer you want. 
So the general advice I would give is that if counting something seems hard, then try to count the opposite. This is one trick of many; combinatorics is a tricky subject, but after learning these little tricks (and a lot of practice) you can get the hang of it!
A: Further to your comments that exactly is implied, I am afraid that the first part of your answer is quite far from correct.
Exactly forces two $L$'s to be together.
Since the lumped two can't be distinguished, we can represent them as one letter, a super $\Bbb{L}$
In $\quad-L-\Bbb{L}-\quad $ the $A$ and $O$ can be placed in $3\cdot 2 = 6$ ways in the gaps,
and $L\;\;and\;\; \Bbb{L}$ can be permuted in two ways,
so by the multiplicative principle the answer = $6\cdot2 = 12$
For the second part, you can use the idea given by Mike Ernest.
