Latest iGate Aptitude Question SOLUTION: How many ways the letters of the word ENGINEERING can be arranged such that NO vowel come together

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(#M40036450) IGATE Logical Reasoning question permutations Keep an EYE

How many ways the letters of the word ENGINEERING can be arranged such that NO vowel come together

Asked In iGate Shalini (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

nitish kumar  (10 years ago)

engineering can be arranged as=
(11!/3!*2!*3!*2!)=277200
now,engineering can be arranged so,that vowels come together=(7!/3!*2!)=4200

so,engineering can be arranged so,that vowels not come together=(277200-4200)=273000 ans

Is this solution Helpfull? Yes (27) | No (7)  

parvathi  (10 years ago)

A)273000,The word enginerring can be written in 277200 ways by the formula
11!/(3!*3!*2!*2!) since there are 3E,3N,2I and 2G,nowthe cases wen the vowels come together are
take all the wovels together as such
NGNRNG(EEEII),take the whole vowel as 1 quantity,so taking the factorial of 7,but not vowel letter has 2G,3N,take (7!/3!*2!)=420,
now the vowels word can be arranged in 5! ways wer there are 3E,2I so take (5!/3!*2!)=10
multiply both we get=420*10=4200
Now subtract the no of ways of arranging the word by wen the vowels comes together, gives the answer=277200-4200=27300

Is this solution Helpfull? Yes (12) | No   

Saraswathy  (10 years ago)

Sorry It might be 8400 ways

No. of vowels = 5
No. of consonants = 6
Let V be a vowel nd C be a consonant
Then possible ways to arrange such that no vowels come together
If starting letter is a vowel, possible combinations are
VCCVCVCVCVC
VCCCVCVCVCV
VCVCCVCVCVC
VCVCCCVCVCV
VCVCVCCVCVC
VCVCVCCCVCV
VCVCVCVCCVC
VCVCVCVCCCV
VCVCVCVCVCC
If starting letter is a consonant, possible combinations are
CCVCVCVCVCV
CVCCVCVCVCV
CVCVCCVCVCV
CVCVCVCCVCV
CVCVCVCVCCV
Total there are 14 possible ways we can arrange so that no vowels ll come together
E=3, N=3, G=2, I=2, R=1
So no. of ways => 14*5!*6!/(3!*3!*2!*2!*1!)
[ 5!=> vowels can be arranged among themselves in 5! ways...6!=> consonants can be arranged among themselves in 6! ways]
= 8400

Ans : 84000

Is this solution Helpfull? Yes (6) | No (25)  

pavan  (10 years ago)

273000 is ans
no of ways aranging engineering is 11!/3!*2!*2!*3!=277200

no of ways to arrange vowels come together is NGNRRNG(EIEEI),EIEEI IS TAKAN AS ONE UNIT THEN 7!/3!*2!=420 WAYS,NO OF WAYS OF ARRANGING VOWELS (EIEEI)IS 5!/3!*2!=10
TOTAL NO OF WAYS ARRANGING ENGINEERING SO THAT VOWELS COME TOGETHER IS 420*10=4200 WAYS

NOW NO OF WAYS OF ARRANGING ENGINEERING SO THAT VOWELS NEVER COM TOGETHER IS =277200-4200=273000

Is this solution Helpfull? Yes (5) | No   

mathuku harani  (10 years ago)

273000 is the best answer

Is this solution Helpfull? Yes (4) | No   

Saraswathy  (10 years ago)

Sorry
Ans : 8400

Is this solution Helpfull? Yes (2) | No (9)  

Gourav Kumar Pandey  (10 years ago)

Guys, you are missing something.
Some of them were correct up to the first part that is:
ENGINEERING can be arranged as=
[{11!/(3!*2!*3!*2!)}]=277200 because of the repetition of the letters: E for 3 times, N for 3 times and I and G for 2 times.

Now, if the vowels are going to come together then the number of ways would be:
[7!/(3!*2!)]*[6!/(3!*2!)]=25200,
the first part as you know that "EIEEI" can be considered as one because we are consolidating them at one place and the repetition of the letters are known to you so we have divided by 3 and 2's factorials.

Here comes the missing part which is to consider the repetition of consonants as well which is consolidated as "NGNRNG" and the repetition is understood to you for what we have divided again 6 letters as 6! by (3!*2!*1!), 3 for N, 2 for g and one for R;

This will give us 25200.

Now the number of ways the letters of the word ENGINEERING can be arranged such that NO vowel come together is:
277200-25200=252000.

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

bhavisha jadav  (8 years ago)

"ENGINEERING" contains 11 letters ,there are repeat alphabt 3E,3N,2G,2I,1R
so, 11!/(3E!)(3N!)(2G!)(2I!)(1R!) ==277200

Is this solution Helpfull? Yes (2) | No   

Saraswathy  (10 years ago)

No. of vowels = 5
No. of consonants = 6
Let V be a vowel nd C be a consonant
Then possible ways to arrange such that no vowels come together
If starting letter is a vowel, possible combinations are
VCCVCVCVCVC
VCVCCVCVCVC
VCVCVCCVCVC
VCVCVCVCCVC
VCVCVCVCVCC
If starting letter is a consonant, possible combinations are
CCVCVCVCVCV
CVCCVCVCVCV
CVCVCCVCVCV
CVCVCVCCVCV
CVCVCVCVCCV
Total there are 10 possible ways we can arrange so that no vowels ll come together
E=3, N=3, G=2, I=2, R=1
So no. of ways => 10*5!*6!/(3!*3!*2!*2!*1!)
[ 5!=> vowels can be arranged among themselves in 5! ways...6!=> consonants can be arranged among themselves in 6! ways]
= 6000

Ans : 6000

Is this solution Helpfull? Yes (1) | No (12)  

Sravani A  (10 years ago)

ans is 27300

Is this solution Helpfull? Yes (1) | No   

B Gurunath  (10 years ago)

Imagine that you arrange the consonants first. There are six consonants which you can arrange in 6!/(3!2!) ways.

Now there are 7 spaces for the 5 vowels to go into but only one vowel can go into each space. So you choose 5 of the 7 available spaces and put a permutation of the vowels into these spaces.

Total number of arrangements with no consectutive vowels =6!/(3!2!)×5!/(3!2!)×(75).

Is this solution Helpfull? Yes | No (1)  

pramit  (8 years ago)

11!-(4200)

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Trinsh Agarwal  (8 years ago)

number of ways to arrange these letters =
11!(3!)(3!)(2!)(2!)=[11×10×9×8×7×6×5×4×3×2×1(3×2×1)(3×2×1)(2×1)(2×1)]=277200

number of arrangements where vowels come together=6! * 5!=86400
then,
No vowel comes together for
277200-86400=190800 (ans.)

Is this solution Helpfull? Yes | No (4)  

Priya  (8 years ago)

vowels are E 3 times, I 2 times
let the word will be NGNRNG(EEEII) having 3 words
total ways= 7!/3! * 2! = 420
Vowel = 5!/3!*2! = 10
answer= 420/10 = 42 ways

Is this solution Helpfull? Yes | No (3)  

POORNIMA  (8 years ago)

engineering can be arranged as=
(11!/3!*2!*3!*2!)=277200
now,engineering can be arranged so,that vowels come together=(7!/3!*2!)=4200
Now, EIEEI has 5 letters in which E occurs 3 times and I occurs two times.

Number of ways of arranging these letters =(5!/3!*2!)
Required number of words = (4200 x 10) = 42000.
so,engineering can be arranged so,that vowels not come together=(277200-42000)=235,200 ans

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