TCS
Company
Numerical Ability
Permutation and Combination
in how many ways can letter of word MOBILE can be arranged so that atleast two consonant remains together
Read Solution (Total 24)
-
- 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 - 8 years agoHelpfull: Yes(26) No(9)
- ans is 576
(total number of arrangements)-(arranging no two consonants to be together)
6!-(4p3*3!)=720-144=576 - 8 years agoHelpfull: Yes(18) No(6)
- 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 - 8 years agoHelpfull: Yes(9) No(4)
- Total number of ways - no consonants to be together
=6! - 3!*4*3*2= 720-144
=576 - 8 years agoHelpfull: Yes(9) No(0)
- 5!2!*3=720
4!3!=144
720+144=864
- 8 years agoHelpfull: Yes(6) No(2)
- 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
- 8 years agoHelpfull: Yes(5) No(2)
- 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. - 8 years agoHelpfull: Yes(5) No(2)
- Answer: 384
Explanation: Two possible cases: Two consonants together + Three consonants together
5! x 2! + 4! x 3! = 384
- 8 years agoHelpfull: Yes(5) No(5)
- the ans ll be 576 only
- 8 years agoHelpfull: Yes(4) No(2)
- they said at least two then we may arrange three together also then it is more than 144
- 8 years agoHelpfull: Yes(3) No(1)
- Guys don't be confused, Think smartly.. You can easily solve it.
Total no.of ways(6!) - NO two cnsnt together(144) = 576 - 8 years agoHelpfull: Yes(2) No(0)
- 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 - 7 years agoHelpfull: Yes(2) No(0)
- 180
there is two case when only 2 consonats(144) take together and other one is when 3 consonants take together(36) total=180 - 8 years agoHelpfull: Yes(1) No(2)
- Answer is 384
2consonent together + 3consonent together
5!*2!+4!*3! - 8 years agoHelpfull: Yes(1) No(2)
- 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. - 8 years agoHelpfull: Yes(1) No(0)
- 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 - 8 years agoHelpfull: Yes(1) No(1)
- 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 - 8 years agoHelpfull: Yes(0) No(3)
- 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 - 8 years agoHelpfull: Yes(0) No(1)
- SRUTHI PUJARI
can u explain ?
how 4p3*3! - 8 years agoHelpfull: Yes(0) No(0)
- ((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 - 8 years agoHelpfull: Yes(0) No(2)
- 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 - 7 years agoHelpfull: Yes(0) No(0)
- 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. - 5 years agoHelpfull: Yes(0) No(0)
- 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 - 2 years agoHelpfull: Yes(0) No(0)
- No(4)Total number of ways - no consonants to be together
=6! - 3!*4*3*2= 720-144
=576 - 1 year agoHelpfull: Yes(0) No(0)
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