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07 Dec 2011, 11:58
Kudos
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Need help with understanding the answer to this problem
How many words can be formed with the letters of the word ‘OMEGA’ when Vowels are never together.
Answer Provided is 84.
My confusion - i tried solving this problem as follows
Firstly, the total number of ways in which the word can be arranged without restrictions is 5! = 120 ways.
Now we have 3 vowels - O,E,A
Scenarios in which the vowels may appear together is when at least 2 vowels appear together.
therefore assuming 2 vowels appearing together as 1 unit. we have 4 words to be arranged this takes into account the scenario where all 3 vowels are placed together
This can be done in 4! ways and the vowels themselvescan be arranged in 3*2 = 6 ways . therefore total ways is 24*6 = 144 ways .this is however greater than the total possible arrangements without restriction
Another alternate I could think of was -
to avoid all the 3 vowels appearing together each vowel needs to be placed in alternate spaces
Vowel1 Consonant Vowel2 Consonant Vowel3
the vowels can be arranged in 3! WAYS and consonants in 2! ways therefore total number of ways = 3!*2! = 24 ways
. Somebody please throw some light. Many thanks in advance
How many words can be formed with the letters of the word ‘OMEGA’ when Vowels are never together.
Answer Provided is 84.
My confusion - i tried solving this problem as follows
Firstly, the total number of ways in which the word can be arranged without restrictions is 5! = 120 ways.
Now we have 3 vowels - O,E,A
Scenarios in which the vowels may appear together is when at least 2 vowels appear together.
therefore assuming 2 vowels appearing together as 1 unit. we have 4 words to be arranged this takes into account the scenario where all 3 vowels are placed together
This can be done in 4! ways and the vowels themselvescan be arranged in 3*2 = 6 ways . therefore total ways is 24*6 = 144 ways .this is however greater than the total possible arrangements without restriction
Another alternate I could think of was -
to avoid all the 3 vowels appearing together each vowel needs to be placed in alternate spaces
Vowel1 Consonant Vowel2 Consonant Vowel3
the vowels can be arranged in 3! WAYS and consonants in 2! ways therefore total number of ways = 3!*2! = 24 ways
. Somebody please throw some light. Many thanks in advance
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07 Dec 2011, 12:50
Kudos
Bookmarks
Hey,
Okay my approach to this question:
Total ways of arranging OMEGA - Total ways of arranging OMEGA so that the vowels are together. ---> This math will provide the answer for the question.
1. Total ways of arranging OMEGA 5 ! ----> 120
2. Total ways of arranging OMEGA so that vowels are together-
To solve this i use MGMAT line method. So according to this, consider the vowels as one element. so considering OEA as 1 element, now its all about arranging 3 elements for which the total ways are 3 ! = 6. Also we must note that the vowel element is made up of 3 vowels which can be re arranged within them. So a total ways of arranging OMEGA with vowels together is 6 * 3! = 36.
So , total ways of arranging OMEGA with vowels not present together = 120 - 36 =84.
Hope it helps.
Okay my approach to this question:
Total ways of arranging OMEGA - Total ways of arranging OMEGA so that the vowels are together. ---> This math will provide the answer for the question.
1. Total ways of arranging OMEGA 5 ! ----> 120
2. Total ways of arranging OMEGA so that vowels are together-
To solve this i use MGMAT line method. So according to this, consider the vowels as one element. so considering OEA as 1 element, now its all about arranging 3 elements for which the total ways are 3 ! = 6. Also we must note that the vowel element is made up of 3 vowels which can be re arranged within them. So a total ways of arranging OMEGA with vowels together is 6 * 3! = 36.
So , total ways of arranging OMEGA with vowels not present together = 120 - 36 =84.
Hope it helps.
07 Dec 2011, 16:06
Kudos
Bookmarks
I think I caught your mistake: 3! x 2! = 12.
My approach is 5! = 120 total.
Now I want to find total ways in which three vowels are together:
VVVCC, CVVVC, CCVVV = 3 possible setups
In each possibility, we get 3! x 2! = 12, Then 12 x 3 (# of setups) = 36.
120 - 36 = 84
My approach is 5! = 120 total.
Now I want to find total ways in which three vowels are together:
VVVCC, CVVVC, CCVVV = 3 possible setups
In each possibility, we get 3! x 2! = 12, Then 12 x 3 (# of setups) = 36.
120 - 36 = 84
