GATE CSE 2020 | Question: 42

Question

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 ______.

Comments

@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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@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 

Answer 1

Answer: 12

Explanation

Here, none of the characters can be at their respective positions in any permutation and we have 2 indistinguishable character- 'L',. Since there are very few characters (5 here) with repeating 'L', we can fix possible positions of 'L' and check. You'll find there are only 3 such possible positions-
1. _ L _ L _
2. _ L _ _ L
3. _ _ _ L L
 

For simplicity, take case 1 and we can conclude that there will be 6 possible strings (3x2x1) which will be-
I L A LC   //NOT ACCEPTABLE SINCE C is at it's correct position
I L C L A
A L C L I
A L I L C    //NOT ACCEPTABLE SINCE C is at it's correct position
C L A L I
C L I L A

This leads us to 4 permutations in one such permutation by fixing L.

Similarly there will be 4 such strings in each of the 3 fixed L cases.

Hence total permutations = 4 + 4 + 4 = 12 (or say 4x3)

Answer 2

(Answer incorrect)

 

Treat the 2 $L$'s in $LILAC$ as distinguishable for the time being. Since it's mentioned in the question

No character appears in its original position

The derangement of $n$ objects is given by 

$n! (1-\frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} . . . . (-1)^n \frac{1}{n!})$

 

Hence derangement of $5$ objects = $5!(1-\frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}+ \frac{1}{4!}- \frac{1}{5!}) = 44$

Now since $L$'s are indistinguishable and there are 2 of them, divide the answer by $2!$ giving $\frac{44}{2} = 22$

Comments

@Prashant.

@d3ba

Can u calculate it manually plz.

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i got 12 , see below answers

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Yes the answer is 12. Tried it using Python

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@d3ba you can apply derangement only when all the elements are distinct. That's not the case here.

Answer 3

Video Solution

https://www.youtube.com/watch?v=X07pj-9KEwM

 

L I L A C

!1 = 0

!2 = 1

!3 = 2(1+0) = 2

!4 = 3(2+1+ = 9

!5 = 4*(9+2) = 44

As $!n = (n-1)[!(n-1)+ !(n-2)]$

Number of ways we can put each alphabet at different place will be !5 i.e. 44

but we have to consider all the cases below

CASE 1: if first L goes in second L’s shoes that means !4 ways are there as I dont care where the second L is

CASE 2: IF second L goes in First L’s shoes that means !4 ways are there as I dont care where the first L is

CASE 3: if First L and Second L swap there place then we have !3 ways of arranging IAC

 

All these cases have to be subtracted from !5

And as we have 2 L’s then we did counting twice, so we will divide by 2!

Ans =  $\frac{!5-!4-!4-!3}{2!}$

which is $\frac{44-9-9-2}{2}=12$

 

 

 

Comments

Then it is very easy,

answer would be 44.

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how bro? please explain.I think there would be intersections amongst.

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If all the elements are distinct then we can use the formula of dearrangement If word is LIZAC 

n=5

5!(1/1! + 1/2! - 1/3! + 1/4! - 1/5! )

Then answer would be 44

Answer 4

First, let's assume that the 2 Ls are distinguishable.

Total number of derangements for 5 items = 5C0*5! – 5C1*4! + 5C2*3! – 5C3*2!+ 5C4*1! – 5 = 24

Now considering the fact that the 2 Ls are the same, the total unique derangements = 24/(2!) = 12

Comments

I guess this approach is somewhat incorrect.
First of all, the value of de-arrangements of 5 distinct elements =  44   and not 24.