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The number of permutations of the letters $a, b, c$ and $d$ such that $b$ does not follow $a,c$ does not follow $b$, and $c$ does not follow $d$, is

  1. $11$
  2. $12$
  3. $13$
  4. $14$
in Combinatory by Veteran (424k points)
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Best answer

Answer: $\mathbf A$

Total number of permutations of $ a, b, c, d = 4!= 24 $

Number of permutations of $ab, c, d = 3!= 6 $
Number of permutations of $a, bc, d = 3!= 6 $
Number of permutations of $a ,b ,cd = 3!= 6 $

Number of permutations of $b \;and\; c = 2!= 2 $
Number of permutations of $c \;and \;d = 2!= 2 $
Number of permutations of $ d \;and \;b = 2! =2 $
Number of permutations of $ b,c \;and \; d = 1! = 1 $

Hence, required permutations are:
$4! - 3! - 3! - 3! + 2! + 2! + 2! - 1! = 11$
 
$\therefore $ option $\mathbf A$ is correct.
by Boss (12.9k points)
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Have you  tried to solve it using inclusion-exclusion principle? Please tell me how do you solve such questions, I am always stuck.

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