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(#M40157302) TCS Numerical Ability question Keep an EYE

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

Asked In TCS gourav (8 years ago)

Unsolved Read Solution (24)

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Read Solution (24) :

Manish Shankar  (8 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  (8 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  (8 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  (8 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  (8 years ago)

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

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

praveen r p  (8 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  (8 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  (8 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  (8 years ago)

the ans ll be 576 only

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

sruthi pujari  (8 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  (8 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  (8 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  (8 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  (8 years ago)

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

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

aruna  (8 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  (8 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  (8 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  (8 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  (8 years ago)

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

Is this solution Helpfull? Yes | No   

guddu  (8 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  (8 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  (5 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  (2 years 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  (1 year ago)

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

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