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Difficult Math Problem #18

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Difficult Math Problem #18

by 800guy » 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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by ajith » Thu Oct 19, 2006 12:26 am
1) Without any restrictions.
As there are 8 distinct Letters -- 8!

2) M must always occur at the third place.
M is in 3rd place
So basically there are 7 letters to arrange in 7 places that can be done in 7! ...


3) All the vowels are together.
There are 3 vowels, consider them as a group
so basically you can arrange C,M,T,R,P and The group of vowels in 6! ways and vowels can be arranged among themselves in 3! ways

So total no of ways 3!*6!

4) All the vowels are never together.

using info from 1 and 3, total No of ways = 8!-6!*3!

5) Vowels occupy the even positions.

Vowels can be arranged in 4P3 ways and the consonants in 4! ways
So total No of ways 4!*4P3
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by rajs.kumar » Thu Oct 19, 2006 4:50 am
[quote="ajith"]5) Vowels occupy the even positions.

Vowels can be arranged in 4P3 ways and the consonants in 4! ways
So total No of ways 4!*4P3[/quote]

ajith - my answers agree with yours except for the 5th one. I will explain my approach let me know what you think.

there are 4 even positions, and 3 vowels. We have to choose 3 places out of 4 = 4C3 = 4C1 = 4.

For each of the 4 choices the vowels can be arranged in 3! ways and the consonants can be arranged in 5! ways.

Total = 4 x 3! x 5!
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by ajith » 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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OA

by 800guy » 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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