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Five letters $A, B, C, D$ and $E$ are arranged so that $A$ and $C$ are always adjacent to each other and $B$ and $E$ are never adjacent to each other. The total number of such arrangements is

  1. $24$
  2. $16$
  3. $12$
  4. $32$
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2 Answers

3 votes
3 votes

 

Consider $AC$ as 1 unit and $D$ as another, total $ 2 unit $

 $AC$  $D$ 

It can be arranged in = $2! * 2! = 4$ ways ( second $ 2! $ because AC can we arranged in to ways i,e AC and CA). $\rightarrow (1) $

 

Now, this arrangement will create three places, i.e $ -- AC -- D--$

Since $B$ and $E$ cannot be adjacent, choose $2$ place out of $3$ = $\binom{3}{2} * 2!$   $\rightarrow (2) $

($2!$ because of BE and EB two arrangement possible).

From $ (1) and (2)$

So total arrangement possible such that $A$ and $C$ are adjacent and $B$ and $E$ are non adjacent

 =  $ 4 * \binom{3}{2} * 2!  $

= $ 24 $

 

 

 

 

 

2 votes
2 votes
$\underline{\textbf{Answer A:}}$

Number of permutations of $$\textbf{A,B,C,D,E} = 5! = 120\;\textbf{ways}$$

Number of permutations of with $\textbf{AC}$ as adjacent $$\mathbf{ \underbrace{AC}_\text { fixed}BDE= 4! 2! = 48\;ways}$$

But in the above result, we need to subtract the case when $\textbf{BE}$ is together $$=\mathbf{ \underbrace{AC}_\text{fixed}\;\;\underbrace{BE}_\text{fixed}D = (3!)2!2! = 24 ways}$$

So, total ways required $$\mathbf{=48-24 = 24\;ways}$$

$\therefore\;\mathbf A$ is the correct option.
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