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If you have the letters LOCAL, how many words can you form

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If you have the letters LOCAL, how many words can you form [#permalink] New post 17 Mar 2005, 09:07
If you have the letters LOCAL, how many words can you form where there is at least one space between the Ls?
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Re: permutation [#permalink] New post 17 Mar 2005, 09:34
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rc1979 wrote:
If you have the letters LOCAL, how many words can you form where there is at least one space between the Ls?


First without considering any conditions, the total arrangements for the word LOCAL is 5!/2 (divided by 2!, since there are two letters same)

Consider the both LL as a group letter, now there will be four letters, which can be arranged in 4! ways.

Hence number of words where there is atleast one space between L's
= [ (5!/2) - 4!] = 36

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 [#permalink] New post 17 Mar 2005, 09:45
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.

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 [#permalink] New post 17 Mar 2005, 09:54
kapslock wrote:
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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 [#permalink] New post 17 Mar 2005, 16:22
Nice explanation Ketan ... Keep it up
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 [#permalink] New post 17 Mar 2005, 17:10
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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 [#permalink] New post 18 Mar 2005, 02:31
(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
  [#permalink] 18 Mar 2005, 02:31
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