Latest TCS Aptitude Question SOLUTION: how many words formed ACCIDENT without changing relative order of vowels??

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(#M40038606) TCS question Keep an EYE

how many words formed ACCIDENT without changing relative order of vowels??

Asked In TCS deepika (9 years ago)

Solved Souradip Nath Read Solution (26)

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Read Solution (26) :

Souradip Nath  (9 years ago)

Vowels: AIE = 1, since order of vowels must not be changed, so 3! should not be multiplied.
Rest: CCDNT = here we have 5 terms, with 1 common term 'C'.

So according to me ans should be :

(5+1)!/2!
= 6!/2!
=360

Is this solution Helpfull? Yes (35) | No (2)  

Ravikant  (9 years ago)

3! * (5!/2!)=360

Is this solution Helpfull? Yes (13) | No (2)  

ranjan kumar  (9 years ago)

count no of vowels and no of consonants at first..3 and 5 but consonants have c 2 times (note)..now take vowels as 1..so now there are 6 words along with consonants..now that can be arranged in 6! ways..but there is c 2 times ..so we will have to divide it by 2! till now the ans is 6!/2! now we should not change the relative order of vowels so we can't interchange them..so there will be only 1 way of putting it..so multiply it with 1! so the ans is 6!/2! * 1! i.e 360

Is this solution Helpfull? Yes (7) | No (2)  

karthikeyank  (9 years ago)

Ans:360
3! *5!/2!=720/2
=360

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

NIVEDA SIVAPRAKASAM  (9 years ago)

Consider all vowels as one letter AIE as X
X C C D N T
6!*3

Is this solution Helpfull? Yes (5) | No (4)  

manoj  (9 years ago)

I dont understand why people are treatin AIE as single unit,despite of question saying relative order of vowels should not be changed
i.e we can have letters between A I E
EX ADNICCET

Is this solution Helpfull? Yes (5) | No   

nmtsaikumar  (9 years ago)

Excluding vowels there are 5 letters and can be written in 5!/2!(as c occurs twice)
And further 3 vowels can be arranged in 3! ways
Total no. of ways=(5!/2!)*3!=360 ways

Is this solution Helpfull? Yes (2) | No   

khushboo  (9 years ago)

count AIE as one term
The rest 5 terms are CCDNT
so ans should be (5+1)!/2!
=6!/2!
=360

Is this solution Helpfull? Yes (2) | No   

vcvb  (9 years ago)

the ans will be=5!/2!

Is this solution Helpfull? Yes (1) | No   

sourav kar  (9 years ago)

5p5*3p3/2p2=360

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

Mayank Sharma  (9 years ago)

8!/(3!.2!)=3360
Consider A,E,I to be blank spaces which are identical and can not rearrange among themselves as they have to appear as A then E then I and C,C,D,N,T can be anywhere in between or at sides.

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

Anjan Oleti  (6 years ago)

Oops! Ans should be 3360 words
8P5 / 2! = 3360

Is this solution Helpfull? Yes (1) | No   

shadhiq batcha  (9 years ago)

8!/2!=8x7x6x5x4x3=2016

Is this solution Helpfull? Yes | No (4)  

karthikeyank  (9 years ago)

Ans:3!*5!
vowels A E I
consonants C C I D N T

Is this solution Helpfull? Yes | No (2)  

shadhiq batcha  (9 years ago)

sorry the above i submitted was wrong

Is this solution Helpfull? Yes | No   

tirupathi reddy  (9 years ago)

Two ways

ACCIDENT
ACCIDETN

Is this solution Helpfull? Yes | No (4)  

nikita  (9 years ago)

order should not be changed thats all... 1st out of 8 remaining arrange in 8p5 ways.. but two c's so divide with 2! and remaining arrange only in 1 way

Is this solution Helpfull? Yes | No   

sandeep singh  (9 years ago)

8-(5+1)=6!/2!=360

Is this solution Helpfull? Yes | No   

ANKIT SHARMA  (9 years ago)

(5^4)*5!/2!

Is this solution Helpfull? Yes | No   

vanitha  (9 years ago)

3!*(6!/2!)=2160

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Ritu Jha  (9 years ago)

but ACIDENTC can also be a soln. the vowels need not be together , cn anyone help me with this

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raviteja  (9 years ago)

i is not a consonant @karthikeyank

Is this solution Helpfull? Yes | No   

N.ANNUSHYA  (9 years ago)

let vowels should be in relative order AEI.
=1!*(6!/2!)
=720/2
=360

Is this solution Helpfull? Yes | No   

saranyadurairaj  (9 years ago)

vowels=AEI should be considered as 1 letter
constants=CCDNT these 5 letters can be arranged in 5 ways and the letter c is repeated so 2!
so totaly 6 letters can be arranged in 6p6/2!=360

Is this solution Helpfull? Yes | No (1)  

pratik kumar  (9 years ago)

AIE = 1 , and rest of the alphabets = 5
so, total no. of words = (5+1)!/2!=360

Is this solution Helpfull? Yes | No (1)  

Anjan Oleti  (6 years ago)

Ans: 840 words.
Calculation: 8P5 / 2!
Explanation:
There are 8 letters, consisting of 5 consonants (2 repeating) and 3 vowels.
The relative order of vowels should not change, i.e. the vowels A, I, E should follow the order among them. In other words, A should come first, followed by I and finally E. But, they need not be together.
So, we could first arrange the 5 consonants in the 8 available places. Then, the remaining 3 places will automatically filled by the vowels in the particular order.
5 Consonants can be filled in 8 places in 8P5 ways.
Since 2 of them repeat, divide by 2!
Hence 8P5 / 2! ways or 840 ways
The arrangement of vowels have to happen in the remaining places in the original order, so they will not create additional ways.
So ans is 840 ways

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