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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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0
Is $A$ the answer??
+1
4!*2 - 3!*2*2= 24
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How you got this??
0
Using some sequence of permutation and how did u got that?
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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

=  $4 * \binom{3}{2} * 2!$

= $24$

Answer $A$

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

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

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

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

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