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Difficult Math Question #66 - Combinations

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Answer

by lalitaroral » Wed Nov 29, 2006 11:21 pm
Total letter count = 8 (COMPUTER)

No of vowels = 3
Available positions for vowels = 4
No of free positions after vowel arrangement = 4-3 = 1

No of consonants= 5
Available positions for consonants= 5 (Inclusive vowel positions)

Lets say positions are=>1 2 3 4 5 6 7 8

3rd place is already fixed for M, leaves 4 consonants to be placed

Lets arrange consonants first==>

position number 1 can be arranged in 4 ways
position number 3 can be arranged in 1 ways
position number 5 can be arranged in 3 ways
position number 7 can be arranged in 2 ways

Leaves one vowel behind, which can be placed on available vowel position

remaining letters(3 vowels+1 consonant) can be arranged as 4! ways for available 4 positions

So the total posible arrangements = 4*1*3*2*4! = 576
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Re: Difficult Math Question #66 - Combinations

by kulksnikhil » Fri Dec 01, 2006 3:47 am
800guy wrote:On how many ways can the letters of the word "COMPUTER" be arranged?
1. M must always occur at the third place
2. Vowels occupy the even positions.
Same logic as above, my answer is

4 * 3 * 1 * 2 * 3 * 1 * 2 * 1 = 144
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OA:

by 800guy » Fri Dec 01, 2006 5:25 pm
OA:

For 1.
7*6*1*5*4*3*2*1=5,040

For 2.) 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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by lalitaroral » Fri Dec 01, 2006 10:21 pm
How abt M's position???

Wont it affect?
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by 800guy » Fri Dec 01, 2006 10:40 pm
lalitaroral wrote:How abt M's position???

Wont it affect?
don't know..that's the only OA provided.
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Re: OA:

by kulksnikhil » Sat Dec 02, 2006 3:11 am
800guy wrote:OA:

For 1.
7*6*1*5*4*3*2*1=5,040

For 2.) 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
aren't 1) and 2) two conditions of the same problem?

Your answer seems like, they are two different questions.
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by Rashmi1804 » Wed Mar 04, 2009 1:33 am
Even if (1) and (2) are two independent problems...
the answers are different..

1) no of way the 7 letters can be arrange in 7 places available....7!


2) No. of even positions available for vowels...4
no. of ways 3 vowels can be arranged in 4 places = 4p3
now, no. of places left for remaining 5 letters = 5
no. of ways these 5 consonents can be arrnaged in 5 places left = 5!
i.e, = 4p3.5! = 480

** WHY IS THE LAST EVEN POSITION NOT AVAILABLE FOR FILLING A VOWEL , AS MENTIONED IN THE OA ?[/u][/i]
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