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Letter combination

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by GmatKiss » Mon Aug 15, 2011 12:25 am
How many words can be formed with the letters of the word 'OMEGA' when,

(i) 'O' and 'A' occupying end places.

(ii) 'E' being always in the middle

(iii) Vowels occupying odd-places

(iv) Vowels being never together.
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by kmittal82 » Mon Aug 15, 2011 12:42 am
(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

(2)
[1] [2] E [3] [4]

[1] can be filled in 4 ways, [2] in 3 ways, [3] in 2 ways , total ways = 4x3x2 = 24

(3)
[V1] [C1] [V2] [C2] [V3] where [V*] is Vowel, [C*] is consonant

V1 can be filled in 3 ways
C1 can be filled in 2 ways
V2 can be filled in 2 ways
C2 can be filled in 1 way
V3 can be filled in 1 way

total ways = 3 x 2 x 2 x 1 x 1 = 12


(4)
Vowels must occupy odd places for this, so same as (3) i.e. 12

OAs please?
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by Options » Mon Aug 15, 2011 5:36 am
kmittal82 wrote:(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

I think you should times another 2! to 6 since O and A can swap places. The qn said at the ends but did not specify which letter at which end.
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by kmittal82 » Mon Aug 15, 2011 5:40 am
Options wrote:
kmittal82 wrote:(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

I think you should times another 2! to 6 since O and A can swap places. The qn said at the ends but did not specify which letter at which end.
ah, thanks for correcting that, agreed, it ought to be 12
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by viv09 » Tue Aug 30, 2011 7:07 pm
Hi. can someone please explain part 4 of the question

I believe for this. we can do the below.
Arrange such that vowels are always together

no of ways the letters can be arranges = 5!

no of ways vowels always together = 2x1x3! = 12
So vowels not together = 120 - 12 = 108..

please correct me if i am wrong

Sorry for the typo earlier.. had to edit
Last edited by viv09 on Tue Aug 30, 2011 7:32 pm, edited 1 time in total.
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by eduardo_cruza » Tue Aug 30, 2011 7:27 pm
i) O and A occupying end places = 12 ways
ii) E being always at the middle = 24 ways
iii) vowels occupying odd places = 12 ways
iv) vowels being never together = 5! - 3!x3 = 102 ways
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by saketk » Wed Aug 31, 2011 11:09 am
viv09 wrote:Hi. can someone please explain part 4 of the question

I believe for this. we can do the below.
Arrange such that vowels are always together

no of ways the letters can be arranges = 5!

no of ways vowels always together = 2x1x3! = 12
So vowels not together = 120 - 12 = 108..

please correct me if i am wrong

Sorry for the typo earlier.. had to edit
Hi-- Your approach is correct but why did you multiply 2 in the part I marked RED?

The answer to 4th part: number of alphabets = 5
so they can be arranged in 5! ways. Now we have a constraint that vowels cannot be together.
So let's find out the cases when vowels are always together-- total vowels 3 (O,E,A) - treat them as 1 letter. so now we have 3 letter M,G & (OEA) -- they can be arranged in 3! ways. Also the 3 vowels can be arranged among themselves in 3! ways.. So total number of ways = 3!*3! = 36

So, our answer will be 5! - 3!*3! = 120- 36 = 84 ways.
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by saketk » Wed Aug 31, 2011 11:12 am
kmittal82 wrote:(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

(2)
[1] [2] E [3] [4]

[1] can be filled in 4 ways, [2] in 3 ways, [3] in 2 ways , total ways = 4x3x2 = 24

(3)
[V1] [C1] [V2] [C2] [V3] where [V*] is Vowel, [C*] is consonant

V1 can be filled in 3 ways
C1 can be filled in 2 ways
V2 can be filled in 2 ways
C2 can be filled in 1 way
V3 can be filled in 1 way

total ways = 3 x 2 x 2 x 1 x 1 = 12


(4)
Vowels must occupy odd places for this, so same as (3) i.e. 12

OAs please?
Hi -- your answer to 4th part is incorrect.. It should be 5!-3!*3! = 84 ways.
why do you think that vowels must occupy odd places?
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by Krk » Wed Aug 31, 2011 11:52 am
Hi -- your answer to 4th part is incorrect.. It should be 5!-3!*3! = 84 ways.
why do you think that vowels must occupy odd places?
Since this is a five letter word 1 2 3 4 5.
If vowels occupy even places, then there are only 2 places where vowels can be used.
Since there are 3 vowels, vowels need to take odd places only.
Hence, answer for 3rd and 4th question has to be same.
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