Latest TCS Aptitude Question SOLUTION: in how many ways can letter of word MOBILE can be arranged so that atleast two consonant remains together

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in how many ways can letter of word MOBILE can be arranged so that atleast two consonant remains together

Asked In TCS gourav (7 years ago)

Unsolved Read Solution (24)

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Read Solution (24) :

Manish Shankar  (7 years ago)

Ans.-144
No. of ways to select 2 consonants from 3=3C2
no. of ways to arrange these two consonants= 3C2*2!
no. of ways to arrange remaining terms=4!
Hence, total no. of ways to arrange= 3C2*2!*4!=144

Is this solution Helpfull? Yes (26) | No (9)  

sruthi pujari  (7 years ago)

ans is 576
(total number of arrangements)-(arranging no two consonants to be together)
6!-(4p3*3!)=720-144=576

Is this solution Helpfull? Yes (18) | No (6)  

LAXMAN KAVITAKE  (7 years ago)

If atleast 2 consonant has to come together
then

1st case MB will come together i.e. MB*O*I*L*E = 5!=120

2nd case MBL will come together i.e. MBL*O*I*E =4!=24

120+24=144

Is this solution Helpfull? Yes (9) | No (4)  

Reeti Mittal  (7 years ago)

Total number of ways - no consonants to be together
=6! - 3!*4*3*2= 720-144
=576

Is this solution Helpfull? Yes (9) | No   

Rohit kumar shaw  (7 years ago)

5!2!*3=720
4!3!=144
720+144=864

Is this solution Helpfull? Yes (6) | No (2)  

praveen r p  (7 years ago)

The word 'MOBILE' has
6
6 letters in which there are
3
3 vowels and
3
3 consonants. Every letter is distinct.

Total number of arrangements possible using the letters of the word 'MOBILE'
=
6
!
=6!

Now find out the different arrangements where no consonants are together.
For this, arrange
3
3 vowels in
3
!
3! ways.
There are
4
4 positions now to place the
3
3 consonants and this can be done in
4P3
4P3 ways.
=> Different arrangements where no consonants are together
=
3
!
×
4P3
=3!×4P3

Therefore, different ways in which the word 'MOBILE' can arranged so that at least 2 consonants come together
=
6
!
−
3
!
×
4P3
=
720
−
(
6
×
24
)
=
720
−
144
=
576
=6!−3!×4P3=720−(6×24)=720−144=576

Is this solution Helpfull? Yes (5) | No (2)  

g.srilekha  (7 years ago)

Case 1:Taking only MB together
MB can be arranged in 5 ways and remining are arranged(along with MB ) in 4! ways.
And also MB can be arranged in 2 ways.
so the no.of ways are 5*4!*2 =>5!*2 =>120.

2nd case: taking MBL
MBL can be arrangd in 4 ways and remining(along with MBL) are arranged in 4*3! ways
MBL can be arranged in 3! ways
so the total no ways are:4!*3! => 144

Adding 2 cases the total no. of way are 144+240=384.

Is this solution Helpfull? Yes (5) | No (2)  

Kritika Goyal  (7 years ago)

Answer: 384
Explanation: Two possible cases: Two consonants together + Three consonants together
5! x 2! + 4! x 3! = 384

Is this solution Helpfull? Yes (5) | No (5)  

dinesh Nag  (7 years ago)

the ans ll be 576 only

Is this solution Helpfull? Yes (4) | No (2)  

sruthi pujari  (7 years ago)

they said at least two then we may arrange three together also then it is more than 144

Is this solution Helpfull? Yes (3) | No (1)  

navi chinnu  (7 years ago)

Guys don't be confused, Think smartly.. You can easily solve it.
Total no.of ways(6!) - NO two cnsnt together(144) = 576

Is this solution Helpfull? Yes (2) | No   

Param  (7 years ago)

When MB is together and is definitely separated from L...then no of ways 4C2*2!*3!*2!=144
This can be done in 3 ways(mb,bl,lm)
So no ways =144*3=432

Now taking MBL together definitely, 4*3!*3!=144

So total no of ways=144+432=576

Is this solution Helpfull? Yes (2) | No   

shubham Gupta  (7 years ago)

180
there is two case when only 2 consonats(144) take together and other one is when 3 consonants take together(36) total=180

Is this solution Helpfull? Yes (1) | No (2)  

Diksha Singh  (7 years ago)

Answer is 384
2consonent together + 3consonent together
5!*2!+4!*3!

Is this solution Helpfull? Yes (1) | No (2)  

aruna  (7 years ago)

total words 6 so 6!
now ques has given that atleast 2 constant and question contain 3 vowels so this vowel can occupy remaining any 4 places so ,
6! - (3!*4P3)= 720-144=576.

Is this solution Helpfull? Yes (1) | No   

ROZINA KHATUN  (7 years ago)

Ans=Total no of arrangements-No of arrangements where no consonants are together

Total no of arrangements= Since all the letters are distinct so that we can arrange MOBILE 6! ways

Total no of arrangements where no consonants are together ,arrangement will look like this

_O_I_E_
So three vowels can be arrange like this in (3P3)3! ways and the remaining Four places can be filled by 3 consonants in 4P3 ways
so total arrangements where no consonants are together is 3!*4P3
so the answer is 6!-(3!*4p3)=576

Is this solution Helpfull? Yes (1) | No (1)  

swetha  (7 years ago)

123456 positions
let us take 12 positions are costent n can be with consonents.They can be arranged in 3c2*2! ways
the other 4 positions arranged in 4! ways.
so 3c2*4!*2!=144 ways

Is this solution Helpfull? Yes | No (3)  

shahid ansari  (7 years ago)

No of arrange way of consonant is=3!+1!=4!
No of arrange way of consonant remain together is=3p2=3!
Total no of way of arrange letter is=4!*3!=144

Is this solution Helpfull? Yes | No (1)  

KARTHIKEYAN  (7 years ago)

SRUTHI PUJARI
can u explain ?
how 4p3*3!

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guddu  (7 years ago)

((total letters - no of consonant + 1 )! / (no of times any vowels are repeat)! ) * (( no of consonant)! / (no of times any consonant are repeat)!
=(6-3+1)! * 3! [ here, no letters are repeated, so no of times repeatation we need not to calculate]
=144

Is this solution Helpfull? Yes | No (2)  

Madhuri Mahato  (7 years ago)

case 1:when 2 consonants are together(i.e. may be MB,BM,LM etc)
no. of way of arrangement among two consonants=3P2
no. of way of arranging this pair along the other letters of the word=5P5

Case 2: :when 3 consonants are together(i.e. may be MBL,BML,BLM etc)
no. of way of arrangement among 3 consonants=3P3
no. of way of arranging these 3 along the other letters of the word=4P4

so, total no. of ways=( 3P2*5P5)+(3P3*4P4)=720+144=864

Is this solution Helpfull? Yes | No   

Rishabh Kundal  (4 years ago)

ans = 576
total no of ways - when no consonant are together
there are total 6 letters so we can arrange them in 6! ways = 6!=720
when no consonant are together then it looks like
_o_i_e_
o, i,e can be arranged in 3! ways
now remaining 4 positions and 3 consonant i.e. = 4P3 ways = 24
so 3!*24= 6*24
=144
so ans = 720-144=576
guys this is the correct answer, plz do not confuse others.

Is this solution Helpfull? Yes | No   

Sayantika  (1 year ago)

in first place we take 2 consonant as fixed and they can interchange its place and remaining 4 place have 4 letters left and they can inter change so the solution is ,

3P2*4P4 = 6*24=144

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Vamsi  (9 Months ago)

No(4)Total number of ways - no consonants to be together
=6! - 3!*4*3*2= 720-144
=576

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