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The number of permutations of the characters in LILAC so that no character appears in its original position, if the two L’s are indistinguishable, is ______.
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Best answer
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64 votes
We have $\text{LILAC}$.

Lets number the positions as $1,2,3,4,5$. Now the two $L$s cannot be placed at position $1$ and $3$ but they can be positioned at $2,4,5$ in $^3C_2= 3$ ways. (since the $Ls$ are indistinguishable)

Now one of $2,4,5$ is vacant. Without loss of generality lets say $2$ is vacant.

Now, if $2$ is vacant we can't palce $I$ there but we can place any of $A$, or $C$, so we have $2$ choices for the position which is left after filling the two $Ls$. Now all of $2,4,5$ are filled.

For the remaining two places $1,3$ we have two characters left and none of them is $L$ so we can place them in $2! = 2 $ ways.

Multiply them all = $^3C_2 * 2 * 2! = 3 * 2 * 2 = 12  \ (ans)$.
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24 votes
24 votes

Problems like these can be solved by taking cases 

19 votes
19 votes

Word is LILAC-

In 1st position A, C and I can come.

Let's start with I - if I comes in 1st position you have 3 choices for 2nd position that are L, A, and C. Now fix L, two choices for 3rd position A and C. If you keep applying this end result would be 

for starting with I

ILACL

ILCLA

IACLL

ICALL

for starting with A

ALICL

ALCIL

ALCLI

ACILL

for starting with C

CLAIL 

CLALI

CLILA

CAILL

So, there are total 12 permutations.

Answer:

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