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Difficult Math Problem #18

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #18

Post Wed Sep 20, 2006 4:58 pm
OA coming after a few responses:

On how many ways can the letters of the word "COMPUTER" be arranged in the following scenarios?

1) Without any restrictions.
2) M must always occur at the third place.
3) All the vowels are together.
4) All the vowels are never together.
5) Vowels occupy the even positions.

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Fri Oct 20, 2006 10:34 am
here's the OA:

1) 8! = 40320
2) 7*6*1*5*4*3*2*1=5,040
3) Considering the 3 vowels as 1 letter, there are five other letters which are consonants C, M, P, T, R
CMPTR (AUE) = 6 letters which can be arranged in 6p6 or 6! Ways
and A, U, E themselves can be arranged in another 3! Ways for a total of 6!*3! Ways

4) Total combinations - all vowels always together
= what u found in 1) - what u found in 3)
= 8! - 6! *3!
5) I think it should be 4 * 720
there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

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ajith Legendary Member
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Post Thu Oct 19, 2006 9:23 pm
Yeah.. I made a mistake ... No of ways in which consonants can be arranged is 5! instead of 4! ... because there are 5 of them ...

so my answer is 4p3*5! .. which coincides with your answer
Thank you for pointin out the mistake

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Fri Oct 20, 2006 10:34 am
here's the OA:

1) 8! = 40320
2) 7*6*1*5*4*3*2*1=5,040
3) Considering the 3 vowels as 1 letter, there are five other letters which are consonants C, M, P, T, R
CMPTR (AUE) = 6 letters which can be arranged in 6p6 or 6! Ways
and A, U, E themselves can be arranged in another 3! Ways for a total of 6!*3! Ways

4) Total combinations - all vowels always together
= what u found in 1) - what u found in 3)
= 8! - 6! *3!
5) I think it should be 4 * 720
there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

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ajith Legendary Member
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Post Thu Oct 19, 2006 9:23 pm
Yeah.. I made a mistake ... No of ways in which consonants can be arranged is 5! instead of 4! ... because there are 5 of them ...

so my answer is 4p3*5! .. which coincides with your answer
Thank you for pointin out the mistake

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