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Combinatorics

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Re: Combinatorics

by Ian Stewart » Fri Jun 27, 2008 7:43 am
kris610 wrote:In how many ways can the letters of the word "COMPUTER" can be arranged so that vowels occupy the even positions ?
None- you don't have enough vowels- at least, if we need to use all eight letters. We have only 3 vowels and 4 'even positions' to fill (2nd, 4th, 6th and 8th). Where is this question from?
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Re: Combinatorics

by atlantic » Fri Jun 27, 2008 8:57 am
Ian Stewart wrote:
kris610 wrote:In how many ways can the letters of the word "COMPUTER" can be arranged so that vowels occupy the even positions ?
None- you don't have enough vowels- at least, if we need to use all eight letters. We have only 3 vowels and 4 'even positions' to fill (2nd, 4th, 6th and 8th). Where is this question from?
Hi Ian,

What about if the question says 'occupy only even positions' instead of 'occupy the even positions'. I believe that the question has a solution if written that way.
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Re: Combinatorics

by Ian Stewart » Fri Jun 27, 2008 10:51 am
atlantic wrote: Hi Ian,

What about if the question says 'occupy only even positions' instead of 'occupy the even positions'. I believe that the question has a solution if written that way.
Well, the first version of the question had a solution also, and that solution was zero. :)

But I think you have correctly identified the intended meaning of the question, and it now becomes a GMAT-like counting problem. I'll post the solution below, but if you haven't attempted it yet, it's worthwhile giving it a shot.

Solution:

We can do this like we do most counting problems: count how many choices we have for each letter, then multiply. Let's look at the odd position first. We have 5 consonants we can use in the odd positions:

1st: 5 choices
3rd: 4 choices
5th: 3 choices
7th: 2 choices

And in the even positions, we can now place the remaining consonant and the three vowels:
2nd: 4 choices
4th: 3 choices
6th: 2 choices
8th: 1 choice

So the answer should be 5!*4!. Luckily all of the letters are different here, so this approach works without any modification to account for the repetition of letters.
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by kris610 » Fri Jun 27, 2008 11:45 am
The question does have enough info, I think. The question says, the vowels should be only in even-numbered positions..This is the way I see it:

You have even-numbered positions, out of which you need to choose 3:4C3
Now, in each of these combinations you can arrange the 3 vowels in 3! ways.
The remaining 5 letters can be arranged in 5! ways.
So, it would be 4C3*3!*5!.

Any comments?
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by Ian Stewart » Fri Jun 27, 2008 2:02 pm
kris610 wrote:The question does have enough info, I think. The question says, the vowels should be only in even-numbered positions..This is the way I see it:

You have even-numbered positions, out of which you need to choose 3:4C3
Now, in each of these combinations you can arrange the 3 vowels in 3! ways.
The remaining 5 letters can be arranged in 5! ways.
So, it would be 4C3*3!*5!.

Any comments?
This is turning into a Sentence Correction question! The question asks how many words can be made "so that vowels occupy the even positions". If you remove the word 'the', all is fine, but the original wording implies that all of the even positions need vowels, which is impossible.

But, atlantic corrected that above, and you've definitely given a correct answer to the intended question. You'll notice that 4C3 = 4, so 4C3*3!*5! = 4*3!*5! = 4!*5!; our answers are identical, even if they look different at first.
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