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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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@Deepak Poonia Sir why this question cannot be solved using derangements i.e 

!n which denotes subfactorial of n??

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@abir_banerjee derangement formula is applicable when all the object and boxes are distinct .

Here the object which is the characters are identical “L” appears twice but boxes which means the positions are distinct .So we cant apply the formula directly.

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edited by

@abir_banerjee

Watch the following “Derangement” Lecture. This question & its variations have been covered in detail.

https://www.youtube.com/watch?v=Ut-eDWWISkE&list=PLIPZ2_p3RNHhTnkf2SkwJU5SG4WEVZaIf&index=2&ab_channel=GOClassesforGATECS 

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8 Answers

64 votes
64 votes
Best answer
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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4 Comments

Are we getting 48 as answer if ls are distinguishable? Or I think it as the word is say LIZAC(same as Ls distinguishable)
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@abdulfahad1 No, it would be the standard derangement problem, so !5 = 44 would be the answer if Ls are distinguishable.
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okay so we have a standard formula for this ..got it...thanks.
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24 votes
24 votes

Problems like these can be solved by taking cases 

1 comment

clear explanation...as compared to other answers
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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.

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3 Comments

When the question ∈ (P-Problem) and not (NP- Problem), most of the times, Brute Force is safer and faster than any knowledge of combinatorics.

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But sadly not creative enough to solve for when word length gets big.
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in some questions not being creative is the key to solve the problem within 2 minutes, cause at the end of the day you just need the answer to be correct
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15 votes
15 votes
Answer:

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