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rc1979
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ketanm
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kapslock
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kapslock
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kapslock
That's what I'd go for too.

5!/2 - 4!.

This is easy since the question read at least one space (alphabet) between the two Ls.

Try the variation of this question of finding the permutations with exactly 1 space (alphabet) between the two Ls.

Wud post my answer in a while.


Two ways of approaching this problem:

1. Consider L X L as a group, where X can be O, C or A.

Thus the variations are (L X L) Y Y where Y Y are the remaining characters.
For one value of X, the possible permutations are 3!. (Two Ys and one LXL).
For all possible three values of X, the possible permutations = 3 * 3! = 18

2. Consider the positions

L _ L _ _ - 3 positions, 3! variations possible
_ L _ L _ - 3 positions, 3! variations possible
_ _ L _ L - 3 positions, 3! variations possible

Total = 3 * 3!
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nocilis
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Nice explanation Ketan ... Keep it up
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vprabhala
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5! number of total words
as we have 2 L's
5!/2.
now we will try to find out in how many ways we can do LL's together. considering them as 1 unit. we can do it 4! ways.

5!/2-4! gives us the number of words without the LL's together..
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(5c2-4c1)*3! or 5!/2! - (4c1*3!) or logic I =>

_ _ _ _ _ => five spaces
L L _ _ _ => 1 way
_ L L _ _ => 1 way
_ _ L L _ => 1 way
_ _ _ L L => 1 way

=> 4 ways where LL are togther => multiply it by 3! => 24 => substract it from the total ways => 5!/2! - 24 = 36 or logic II =>

_ _ _ _ _ => five spaces
L _ _ _ L => 1 way
L _ _ L _ => 1 way
L _ L _ _ => 1 way
_ L _ L _ => 1 way
_ L _ _ L => 1 way
_ _ _ L L => 1 way

=> 6 ways where LL are NOT together => multiply it by 3! => 36

many ways to rome :-D



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