Probability question

[GauravShah]

Originally Posted by QuintEssence

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

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.

[pendyal]

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..

[TheAkshaT]

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

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!

[QuintEssence]

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!

[ritika_j]

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

[supervish]

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

[pendyal]

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..

[supervish]

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

[ritika_j]

hey ppl, make it a little simple..... i didnt get ur explanations.

[pendyal]

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