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  # How many 7 letter words can be constructed using 26 letters
  
  Author Message
  Intern
  Joined: 26 Mar 2006
  Posts: 27
  How many 7 letter words can be constructed using 26 letters [#permalink]
  
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  12 Apr 2006, 20:09
  This topic is locked. If you want to discuss this question please re-post it in the respective forum.
  
  How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)
  
  a) (21^4)*(5^3)
  b) (21^4)*(5^3)*7!
  c) {(21^4)*(5^3)*7!}/4!*3!
  d) 21C4*5C3*7!
  e) None of these
  VP
  Joined: 29 Dec 2005
  Posts: 1341
  Re: P&C - 7 letter word [#permalink]
  
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  12 Apr 2006, 21:28
  chillpill wrote:
  How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)
  
  a) (21^4)*(5^3)
  b) (21^4)*(5^3)*7!
  c) {(21^4)*(5^3)*7!}/4!*3!
  d) 21C4*5C3*7!
  e) None of these
  
  total = 26
  vowels (a, e, i, o, and u) = 5
  consonants = 21
  
  no of ways 3 vowels can be selected from 5 vowels = 5c3
  no of ways 4 (7-3) consonants can be selected from 21 consonants= 21c4
  no of ways 7 words can be arranged = 7!
  so multiply all then its
  
  (D) 21C4*5C3*7!
  Director
  Joined: 10 Oct 2005
  Posts: 718
  Re: P&C - 7 letter word [#permalink]
  
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  12 Apr 2006, 21:39
  Professor wrote:
  chillpill wrote:
  How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)
  
  a) (21^4)*(5^3)
  b) (21^4)*(5^3)*7!
  c) {(21^4)*(5^3)*7!}/4!*3!
  d) 21C4*5C3*7!
  e) None of these
  
  total = 26
  vowels (a, e, i, o, and u) = 5
  consonants = 21
  
  no of ways 3 vowels can be selected from 5 vowels = 5c3
  no of ways 4 (7-3) consonants can be selected from 21 consonants= 21c4
  no of ways 7 words can be arranged = 7!
  so multiply all then its
  
  (D) 21C4*5C3*7!
  
  10x you are really an El Professore)))
  _________________
  
  IE IMBA 2010
  
  Director
  Joined: 13 Nov 2003
  Posts: 789
  Location: BULGARIA
  
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  13 Apr 2006, 00:29
  Think that since repetitions are allowed it should be B)
  Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)
  Director
  Joined: 24 Oct 2005
  Posts: 659
  Location: London
  
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  13 Apr 2006, 03:17
  The 7 letter word can be
  
  V V V C C C C which can be arranged in 7! ways.
  
  Each V can be any of the 5 vowels.
  Each C can be any of the 21 Consonants - since repetition is allowed.
  
  (5*5*5 * 21*21*21*21 ) * 7!
  
  Ans = B
  Manager
  Joined: 09 Feb 2006
  Posts: 129
  Location: New York, NY
  
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  13 Apr 2006, 07:01
  I thought the words do have to be different. This would give us A, correct?
  VP
  Joined: 28 Mar 2006
  Posts: 1369
  
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  13 Apr 2006, 15:25
  BG is right...D is correct
  VP
  Joined: 29 Dec 2005
  Posts: 1341
  
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  13 Apr 2006, 17:56
  BG wrote:
  Think that since repetitions are allowed it should be B)
  Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)
  
  you are right. i didnot cinsider "repititon."
  Senior Manager
  Joined: 24 Jan 2006
  Posts: 252
  
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  14 Apr 2006, 08:17
  
  (5*5*5 * 21*21*21*21) * 7!
  Senior Manager
  Joined: 08 Sep 2004
  Posts: 257
  Location: New York City, USA
  
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  14 Apr 2006, 20:56
  Professor wrote:
  BG wrote:
  Think that since repetitions are allowed it should be B)
  Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)
  
  you are right. i didnot cinsider "repititon."
  
  But even if you consider repetition, A should not be the answer. I think C is the correct Answer.
  
  - Vipin
  14 Apr 2006, 20:56
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