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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 (430k points)
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1 Answer

+3 votes
Best answer

$\mathbf{\underline{Answer: A}}$

$\underline{\mathbf{Explanation:}}$

Total number of permutations of $\color{darkorange}{\mathbf{a, b, c, d}} = 4!= 24 $

Number of permutations of $\color{blue}{\mathbf{ab, c, d} }= 3!= 6 $
Number of permutations of $\color{blue}{\mathbf{a, bc, d}} = 3!= 6 $
Number of permutations of $\color{blue}{\mathbf{a ,b ,cd} }= 3!= 6 $

Number of permutations of $\color{green}{\mathbf b} \;\text{and}\; \color{green}{\mathbf c} = 2!= 2 $
Number of permutations of $\color{green}{\mathbf c} \;\text{and} \;\color{green}{\mathbf d} = 2!= 2 $
Number of permutations of $ \color{green}{\mathbf d} \;\text{and} \;\color{green}{\mathbf b}= 2! =2 $

Number of permutations of $ \color{green}{\mathbf{b,c}} \;\text{and} \; \color{green}{\mathbf d} = 1! = 1 $

Hence, required permutations are:

$\mathbf{4! - 3! - 3! - 3! + 2! + 2! + 2! - 1! = \underline{11}}$
 
$\therefore $ Option $\mathbf A$ is the correct answer.
by Boss (18.9k points)
edited by
0
Have you  tried to solve it using inclusion-exclusion principle? Please tell me how do you solve such questions, I am always stuck.
0
Just take a simple test case and analyse it in and out, this is the best way to answer such questions easily.

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