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by chaitanyareddy » Tue Nov 09, 2010 8:21 pm
Hi please answer the following three questions :

1) In how many ways can the letters of the word COMPUTER can be arranges such that vowels

occupy even number.

2) In how many ways can 5 people be seated in a circular table with one should not have the

same neighbour in any two arrangements.

3) There are 6 questions in a question paper . How many ways can a student solve one or more

questions.
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by diebeatsthegmat » Tue Nov 09, 2010 8:42 pm
chaitanyareddy wrote:Hi please answer the following three questions :

1) In how many ways can the letters of the word COMPUTER can be arranges such that vowels

occupy even number.

2) In how many ways can 5 people be seated in a circular table with one should not have the

same neighbour in any two arrangements.

3) There are 6 questions in a question paper . How many ways can a student solve one or more

questions.
i dont understand the questention 2 or i didnt try to but maybe later
1/ COMPUTER there are 8 numbers from 1 2 3 4 5 6 7 8
we must rearrangle the letters in order O, E, U be in even number ( 2,4,6,8). because there are 4 place for 3 number we will have 4!/3!=4 ways to choose places
3 vowels letters =3! ways
since we choose 3 places for 3 letters there will be 5 places for 5 letters left =5!
total = 5!*3!*4

3...
select the combinations that the students can solve 1 question from 1 to 6 = 6
2 questions: 6!/2!4!=15
3 questions : 6!/3!3!=20
4 questions=15
5 questions=6
all 6 =1
total = 6+15+20+15+6+1
are those right answers?
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by beat_gmat_09 » Tue Nov 09, 2010 10:23 pm
chaitanyareddy wrote:Hi please answer the following three questions :

1) In how many ways can the letters of the word COMPUTER can be arranges such that vowels

occupy even number.

2) In how many ways can 5 people be seated in a circular table with one should not have the

same neighbour in any two arrangements.

3) There are 6 questions in a question paper . How many ways can a student solve one or more

questions.
1) 3 vowels can be put in 4 places. No. of ways - P(4,3) = 4!
remaining 5 letters can be arranged in 5! ways.
Therefore 5! * 4! ways of arranging.

2) Circular arrangements given by (n-1)!= (5-1)!=4!

3) C(6,1)+C(6,2)+C(6,3)+C(6,4)+C(6,5)+C(6,6) = 48.
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by chaitanyareddy » Tue Nov 09, 2010 11:11 pm
@beat_gmat_09,

I think the answer to the third question is 63.

and for the second question , the general formula which you gave ( n-1)! is applicable for normal cases of circular arrangement. This is a different case.
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by Rahul@gurome » Thu Nov 11, 2010 12:06 am
2) In how many ways can 5 people be seated in a circular table with one should not have the
same neighbour in any two arrangements.

The number of ways of arranging 5 people in a circular way is (5-1)! = 4!.
If a person P has neighbors A and B, the two ways in which they can sit is APB and BPA.
So for a arrangement if a person has two particular neighbors, there will be one more arrangement with the same two neighbors.
Hence, for each person, with same neighbors, we have 2 such arrangements in 4!.
So we can say that there are 4!/2 or 24/2 = 12 arrangements where no person has same two neighbors
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