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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 ________ . Note - This question was Numerical Type.
  • a)
    12
  • b)
    10
  • c)
    8
  • d)
    15
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The number of permutations of the characters in LILAC so that no chara...
There are 3 choices for the first slot, and then 2 for the third slot. That leaves one letter out of I,A,C unchosen and there are 2 slots that one might occupy. After that, the L′s must go in the 2 unfilled slots. Hence the answer is,
3×2×1×2 = 12 
Option (A) is correct.
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Most Upvoted Answer
The number of permutations of the characters in LILAC so that no chara...
There are 3 choices for the first slot, and then 2 for the third slot. That leaves one letter out of I,A,C unchosen and there are 2 slots that one might occupy. After that, the L′s must go in the 2 unfilled slots. Hence the answer is,
3×2×1×2 = 12 
Option (A) is correct.
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Community Answer
The number of permutations of the characters in LILAC so that no chara...
There are two possible cases to consider when counting the number of permutations of the characters in LILAC so that no character appears in its original position:

1. The two L's are adjacent:
In this case, the two L's can only be in two positions - either in the first two positions (LLIAC) or in the last two positions (ILALC). For the remaining three characters (I, A, and C), there are 3! = 6 permutations. So, in total, there are 2 * 6 = 12 permutations in this case.

2. The two L's are not adjacent:
In this case, the two L's cannot be in the first or last positions, as they would be in their original positions. This leaves us with three possible positions for the two L's: in the second and fourth positions (ILALC), in the second and fifth positions (IALLC), or in the third and fifth positions (IALCL). For the remaining three characters (I, A, and C), there are 3! = 6 permutations. So, in total, there are 3 * 6 = 18 permutations in this case.

Therefore, the total number of permutations of the characters in LILAC so that no character appears in its original position is 12 + 18 = 30.
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