The Gateway to Computer Science Excellence
+1 vote
26 views

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. $32$
in Combinatory by Veteran (424k points)
recategorized by | 26 views
0
Is $A$ the answer??
+1
4!*2 - 3!*2*2= 24
0
How you got this??
0
Using some sequence of permutation and how did u got that?
0
Yes, the answer is correct.

2 Answers

0 votes
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.
by Boss (12.9k points)
edited by
0 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 $

 

 

 

 

 

by Junior (657 points)
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,648 questions
56,430 answers
195,211 comments
99,927 users