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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 19 Jan 2005, 20:15
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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 [#permalink] New post 20 Jan 2005, 04:29
How many words : 5!/2! = 60

Number of words with the 2 Ls next to each other : 4*3! = 24

My answer : 60 - 24 = 36
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 [#permalink] New post 20 Jan 2005, 06:11
Same approach as twixt and also got 36
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 [#permalink] New post 20 Jan 2005, 07:08
twixt wrote:
How many words : 5!/2! = 60

Number of words with the 2 Ls next to each other : 4*3! = 24

My answer : 60 - 24 = 36

:roll:

Last edited by vprabhala on 20 Jan 2005, 12:39, edited 1 time in total.
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 [#permalink] New post 20 Jan 2005, 07:45
vprabhala wrote:
twixt wrote:
How many words : 5!/2! = 60

Number of words with the 2 Ls next to each other : 4*3! = 24

My answer : 60 - 24 = 36

5! is the number of letters.. and 2! we use cauz for a minimum word we need atleast 2 words is that it?

No, it's because letter "L" repeats 2 times
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 [#permalink] New post 20 Jan 2005, 15:45
LOCAL

Case 1
3 Spaces L x x x L the 3 OCA can be arranged 3! = 6

Case 2

2 spaces L x x L x or x L x x L can 3C2 * 3C2 ways = 9

Case 3

1 Space will be 3 ways

total number of ways will be 6 * 9 * 3 = 162 ways

RC can u please post the OA
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 [#permalink] New post 20 Jan 2005, 16:31
rxs0005 wrote:
LOCAL

Case 1
3 Spaces L x x x L the 3 OCA can be arranged 3! = 6

Case 2

2 spaces L x x L x or x L x x L can 3C2 * 3C2 ways = 9

Case 3

1 Space will be 3 ways

total number of ways will be 6 * 9 * 3 = 162 ways

RC can u please post the OA

rxs, I see you approached this problem the long way with favorable outcome approach. I agree with case 1 but for cases 2 and 3, there is a mistake in your reasoning

case2
LxxLx
xLxxL
For each of the above 2, you have 3! ways of arranging 3 other letters. So you have 3!*2 = 12

case3
LxLxx
xLxLx
xxLxL
Once again, for each of these possible ways of positioning "L", you have 3! ways of doing so. So it is 3!*3 = 18

So you have total favorable outcomes = case(1+2+3) = 6+12+18 = 36
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 [#permalink] New post 20 Jan 2005, 16:36
hey paul

thanks for the explantaion i mixed up counting with combinations

i get it now

rxs
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 [#permalink] New post 20 Jan 2005, 16:49
rxs0005 wrote:
LOCAL

Case 1
3 Spaces L x x x L the 3 OCA can be arranged 3! = 6

Case 2

2 spaces L x x L x or x L x x L can 3C2 * 3C2 ways = 9

Case 3

1 Space will be 3 ways

total number of ways will be 6 * 9 * 3 = 162 ways

RC can u please post the OA


rxs, case 1 is good. But there is a problem in your case 2 and case 3.

For case 2:
For L _ _ L _ : You will have 3! ways, not 3C2
LOCLA;
LCOLA
LAOLC
LOALC
LACLO
LCALO

Similarly, for _ L _ _ L : You will have another 3! ways.

Case 3:
It should not be just 3 ways.
You will have:
L _ L _ _ : 3! ways for this arrangement
_ L _ L _ : 3! ways for this arrangement
_ _ L _ L : 3! ways for this arrangement

In the end you need to add all the ways not multiply.
So you get 6(3!) = 6*6 = 36
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 [#permalink] New post 20 Jan 2005, 16:50
:oops: oops, did not realize Paul had already explained.
  [#permalink] 20 Jan 2005, 16:50
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