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GauravShah
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08-12-2004, 03:23 PM

Quote:
Originally Posted by QuintEssence
Quote:
Originally Posted by GauravShah
Quote:
Originally Posted by ritika_j
In how many ways can the letter of the word "daughter" be arranged such that no vowels are together?

answer this question with explanation
Not sure if my approach is correct still here is my solution.....

Lets place this first V-C-V-C-V...here three vowels can be arrange in 3! = 6 ways and consonants in 5P2 = 5.4 = 20 ways...

hence total = 6*20 = 120 ways...

Now consider the spaces between these that is...

_ - V - _ - C - _ - V - _ - C - _ - V - _

In this stepp we fill these spces will the remaining consonants i.e. 3

This can be done in 6C3 * 3! ways = 20*6 = 120 ways

Hence total no. of ways = 120*120 = 14400 ways

If I missed something please reply...

Gaurav.
well champ, u did miss something..u see "daughter" has 3 vowels nd 5 consonants..so its not necessary to have a vc-vc or -v-c sort of repetitive relatonship..actually the solution is much simpler

step1: calculate all possible permutations of "daughter"= 8!

step2: calculate all cases in which three vowels are always together=3!x6!

step3: calculate cases of two vowels being together..this is slightly tricky, so a diagram follows

_ _ _ _ _ _ _ _
1 2 3 4 5 6 7 8 lets consider two vowels at 1-2 position..3C2=3 arrangements
=> rest arrangements can be made in 6! ways

out of these 6! arrangements 5! arrangements will have a vowel at place 3

so we have 3x(6!-5!) ways

now this 1-2 group can be taken in 7 ways

so arrangements for two vowels together are= 7x3x(6!-5!)

hence, we have total arrangements= 8!- [3!x6! + 7x3x(6!-5!)]

I realized I had made an error and had prepared another long solution to correct it

Did get u when u said this

Quote:
u see "daughter" has 3 vowels nd 5 consonants..so its not necessary to have a vc-vc or -v-c sort of repetitive relatonship
Anywez your solutions seems perfect....nice one

Gaurav.


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Permutations - 08-12-2004, 04:28 PM

i am coming up with a different answer.cud u tell me if it is right.

first arrange all the 5 consonants in 5! ways.
now the 3 vowels must go in between the consonantsor at the extremes.so there are 6 possible places for the consonants to occupy.
we can choose 3 places from the six in 6P3 ways i.e.120 ways.

so the total possible no. of ways are 5!*120=14400.

bye..


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08-12-2004, 05:51 PM

Quote:
Originally Posted by QuintEssence
step1: calculate all possible permutations of "daughter"= 8!

step2: calculate all cases in which three vowels are always together=3!x6!

step3: calculate cases of two vowels being together..this is slightly tricky, so a diagram follows

_ _ _ _ _ _ _ _
1 2 3 4 5 6 7 8 lets consider two vowels at 1-2 position..3C2=3 arrangements
=> rest arrangements can be made in 6! ways
that'll be 6! * 2! ways....as those 2 vowels can also be interchanged
Quote:
out of these 6! arrangements 5! arrangements will have a vowel at place 3

so we have 3x(6!-5!) ways

now this 1-2 group can be taken in 7 ways

so arrangements for two vowels together are= 7x3x(6!-5!)

hence, we have total arrangements= 8!- [3!x6! + 7x3x(6!-5!)]
Multiply 2! everywhere

-Akshat

PS: pendyal ur solution too seems correct!
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08-12-2004, 06:55 PM

Quote:
that'll be 6! * 2! ways....as those 2 vowels can also be interchanged

Multiply 2! everywhere
-Akshat
u r right..i do need to multiply by 2!
   
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09-12-2004, 05:10 PM

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
   
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09-12-2004, 05:15 PM

Quote:
Originally Posted by ritika_j
Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
Sample space = 64C2

Seven unique adjacent square sets in each row and each column.

Thats 7*(8 rows + 8 columns) = 112.

P(ur event) = 112 / 64C2


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09-12-2004, 06:22 PM

Quote:
Originally Posted by ritika_j
Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
total no. of squares on a chess board=1^2+2^2+3^2+...+8^2=204

sample space = C(204,2)

now assuming that 2 squares can have a side common only if they are of the same dimensions:

no squares of side>4 have a side in common.
no. of possibilities of 2 squares of side 4 having a common side=4
no. of possibilities of 2 squares of side 3 having a common side=36
no. of possibilities of 2 squares of side 2 having a common side=70
no. of possibilities of 2 squares of side 1 having a common side=112

so the total no. of favourable cases is 112+70+36+4=222

hence probability is 222/C(204,2)

hope this is right.
bye..


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09-12-2004, 06:28 PM

Quote:
Originally Posted by pendyal
total no. of squares on a chess board=1^2+2^2+3^2+...+8^2=204

sample space = C(204,2)
Gee... I'm a dope.

How could I say 64 squares ?

However if you modify the Q to say "Two homogenous squares " I'm right


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10-12-2004, 10:18 AM

hey ppl, make it a little simple..... i didnt get ur explanations.
   
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10-12-2004, 10:49 AM

i just took up a paper and a pencil and counted out the no. of possibilities where two square of same size have a side in common.

kinda grude but cudnt think of a better way

any way hard work always pays off,doesnt it?

bye..

p.s: is the answer correct


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