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MANAGEMENT

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by pepeprepa » Wed Jul 23, 2008 8:09 am
I have 5400
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Re: MANAGEMENT

by egybs » Wed Jul 23, 2008 8:19 am
parallel_chase wrote:In how many ways the letters of the word MANAGEMENT can be arranged with vowels in even positions.

OA coming after few discussions.
5!/(2!*2!) = 5*3*2 = 30.
6!/(2!*2!) = 6*5*3*2 =180

180*30 = 5400

Not 100% confident with this though. I think it should work because when you pick the locations of the consonants, there is one extra consonant that will fill the open even location... But if there had been more than one open even spot, we would have to do another round of multiplication.
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by parallel_chase » Wed Jul 23, 2008 8:58 am
There are 4 consonants and 2 vowels. The vowels are in repetition twice.

Vowels can be arranged in 5 even positions = 4! * 5C4

Remaining 6 positions can can be filled in 6! ways

M A E N repeated twice.

5!*6!/ (2!*2!*2!*2!) = 5400.

Thanks for the discussions
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by sudhir3127 » Wed Jul 23, 2008 9:16 am
thanks parallel,,,

was just wondering that there are only 4 vowels ( AA EE)
they can arrange themselves 4 even places in 4p3 ways
5 consonants can arrange in 5 odd places in 5! ways
1 consonant can arrange itself in 1 even place in 6 ways

hence it shud be 4!*6!/2!*2!*2!*2!= 2160?

let me know where i am wrong?
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by pepeprepa » Wed Jul 23, 2008 11:10 am
(In your calculation, it is not 2160)
I think you forget that there are 5 even possible places.
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by sudhir3127 » Wed Jul 23, 2008 11:18 am
though there are 5 even places but only 4 vowels.. hence i used that one blank even place for 6 consonant as there are only 5 odd places for 6 consonants,,
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by pepeprepa » Wed Jul 23, 2008 1:03 pm
I think you do not count all the different even places which can exist
Here are the vowel places.
_ X _ X _ X _ X _ X

(I dont write the consonants in the following examples it is useless)
I think you count only 1 out of 5 possibilities because of the place where there is no vowel: for example you only count AAEE_ but there are also AAE_E, AA_EE, A_AEE, _AAEE --> 5 way
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