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

• 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
• 10 years agoHelpfull: Yes(27) No(7)
• 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
• 10 years agoHelpfull: Yes(12) No(0)
• 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
  
  
• 10 years agoHelpfull: Yes(6) No(25)
• 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
• 10 years agoHelpfull: Yes(5) No(0)
• 273000 is the best answer
• 10 years agoHelpfull: Yes(4) No(0)
• Sorry
  Ans : 8400
• 10 years agoHelpfull: Yes(2) No(9)
• 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.
• 10 years agoHelpfull: Yes(2) No(7)
• "ENGINEERING" contains 11 letters ,there are repeat alphabt 3E,3N,2G,2I,1R
  so, 11!/(3E!)(3N!)(2G!)(2I!)(1R!) ==277200
  
• 8 years agoHelpfull: Yes(2) No(0)
• 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
  
• 10 years agoHelpfull: Yes(1) No(12)
• ans is 27300
• 10 years agoHelpfull: Yes(1) No(0)
• 
  
  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).
  
• 10 years agoHelpfull: Yes(0) No(1)
• 11!-(4200)
• 8 years agoHelpfull: Yes(0) No(1)
• 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.)
• 8 years agoHelpfull: Yes(0) No(4)
• 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
  
• 8 years agoHelpfull: Yes(0) No(3)
• 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
• 8 years agoHelpfull: Yes(0) No(0)