Official Quant thread for CAT 2013

scrabbler · February 21, 2013, 11:26am

@IIMAIM said:
No,It is like this1m and 1w can be chosen among 15m and 15w is 15c1*15c1=225again 1m and 1w can be chosen among 14m and 14wis 14c1*14c1=196.again 1m and 1w can be chosen among 13m and 13wis 13c1*13c1=169..similarly 1m and 1w can be chosen among 1m and 1w is 1c1*1c1=1 sum of squares till 15 which is ÎŁn^2 (where n=15) = 1240

You are ordering the couples? We just have to separate them right, nothing is said about the order of the couples being important...so 15! seems logical...

regards
scrabbler

insane.vodka · February 21, 2013, 11:36am

@jain4444 said:
the digits 1, 2 , 3, ....,9are written at random to form a Nine-digit number..... The probability that it is divisible by "11" is

It will be 2 cases..
one sum of alternate digits should be equal
second difference between them 11k multiple
as 1 to 9 numbers add up to 45 first case is not possible
in second case
sum of 4 digits will be 17/28
or sum of 5 digits will be 28/17
now what to do next

logrhythm · February 21, 2013, 11:41am

@jain4444 said:
The no of ways of dividing 15 men and women into 15 couples each consisting of man and woman is

Is it 15!??

choose 1 man in 15c1 way...now since we need to form a couple the female wld automatically be chosen..

similarly next man in 14c1..

13c1...1c1

hence 15! ways??

scrabbler · February 21, 2013, 11:44am

@jain4444 said:
the digits 1, 2 , 3, ....,9are written at random to form a Nine-digit number..... The probability that it is divisible by "11" is

11 / 126?

Pata nahin kyun yeh lag raha hai.

Listed cases, got 11 sets whose sum satisfies the condition, and so 11 * 5! * 4! out of 9! OA?

@insane.vodka bhai zara dekh lena if this matches whatever you got. Me off now...

regards
scrabbler

scrabbler · February 21, 2013, 11:47am

@Ibanez said:
PS: If we are asked not to arrange the order then there is only ONE way we can group 15men with 15 women since each gets one.

Really? If only it were that simple... *damn real life*

regards
scrabbler

logrhythm · February 21, 2013, 11:50am

@jain4444 said:
the digits 1, 2 , 3, ....,9are written at random to form a Nine-digit number..... The probability that it is divisible by "11" is

a+p+b+q+c+r+d+s+e

(a+b+c+d+e) - (p+q+r+s) = 0 or 11k

too cumbersome to be done in office... 😞

insane.vodka · February 21, 2013, 11:51am

@mailtoankit said:
@insane.vodka haan bhai...aagaye quant thread pe...back to square one??

no SB :nono:

ibanez · February 21, 2013, 11:52am

@scrabbler said:
Really? If only it were that simple... *damn real life*regardsscrabbler

My bad. Forgot to mention "as per the given condition, each one gets one" hahaha

viewpt · February 21, 2013, 11:53am

@Logrhythm said:
a+p+b+q+c+r+d+s+e (a+b+c+d+e) - (p+q+r+s) = 0 or 11ktoo cumbersome to be done in office...

I too agree..mujhe toh abhi abhi kisi dusri company wale ka interview k liye call aaya mere boss k samne..:)
@jain4444 bhai bauncer kara rahey ho ek k baad ek..

insane.vodka · February 21, 2013, 11:53am

@Logrhythm said:
a+p+b+q+c+r+d+s+e (a+b+c+d+e) - (p+q+r+s) = 0 or 11ktoo cumbersome to be done in office...

digits add up to 45 so case with 0 can be neglected
in 11k case difference cannot go to 22
and to get difference 11 there is only one possibility of
17 28
How to go beyond this...
too weak at Probability :splat:

logrhythm · February 21, 2013, 11:55am

@viewpt said:
I too agree..mujhe toh abhi abhi kisi dusri company wale ka interview k liye call aaya mere boss k samne..

Congrats... :D

@insane.vodka said:
digits add up to 45 so case with 0 can be neglectedin 11k case difference cannot go to 22and to get difference 11 there is only one possibility of 17 28How to go beyond this...too weak at Probability

It is not probability after this...it is number system...

total cases are 9!

and i haven't worked on the cases where the divisibility gets satisfied so can't comment on that...

insane.vodka · February 21, 2013, 11:58am

\

@Logrhythm said:
Congrats... It is not probability after this...it is number system...total cases are 9! and i haven't worked on the cases where the divisibility gets satisfied so can't comment on that...

bhai, when you solve this...tag me

:p

dexian · February 21, 2013, 12:01pm

(9 5 2 1 ) ( 9 4 3 1 ) ( 8 6 2 1) ( 8 5 3 1) (8 4 3 2 ) (7 6 3 1) (7 5 3 2) (6 5 4 2)

i m getting only 8 sets of 4 digit adding up to 17.... rest 5 will add up to 28..

@scrabbler how r u getting 11 sets...

plzz xplain.. which r d ones i m missing..

insane.vodka · February 21, 2013, 12:04pm

@scrabbler said:
11 / 126?Pata nahin kyun yeh lag raha hai.Listed cases, got 11 sets whose sum satisfies the condition, and so 11 * 5! * 4! out of 9! [email protected] bhai zara dekh lena if this matches whatever you got. Me off now...regardsscrabbler

their arrangement is different so we will have to *2 for it??

scrabbler · February 21, 2013, 12:06pm

@Dexian said:
(9 5 2 1 ) ( 9 4 3 1 ) ( 8 6 2 1) ( 8 5 3 1) (8 4 3 2 ) (7 6 3 1) (7 5 3 2) (6 5 4 2)i m getting only 8 sets of 4 digit adding up to 17.... rest 5 will add up to 28..@scrabbler how r u getting 11 sets...plzz xplain.. which r d ones i m missing..

(7 5 4 1) bhi ek hai.

Also if the 4 add up to 28, rest will add to 17, woh bhi chalega na...so (9, 8, 7, 4), (9, 8, 6, 5) bhi aayenge...have I missed anything?

regards
scrabbler

scrabbler · February 21, 2013, 12:08pm

@insane.vodka said:
their arrangement is different so we will have to *2 for it??

No, the 5 digits are in the 1, 3, 5, 7, 9th position only, the others in 2, 4, 6, 8. Odd no of digits na?

regards
scrabbler

jain4444 · February 21, 2013, 12:10pm

The total number of Integral solutions of u²v²w²x²y =648000

dexian · February 21, 2013, 12:11pm

@scrabbler

correct hai sir,

i missed these three sets..

so 11 * 5! * 4! / 9!.....

is there any other way out....

insane.vodka · February 21, 2013, 12:15pm

@jain4444 said:
The total number of Integral solutions of u²v²w²x²y =648000

90?? :splat:

jain4444 · February 21, 2013, 12:15pm

@scrabbler said:
11 / 126?Pata nahin kyun yeh lag raha hai.Listed cases, got 11 sets whose sum satisfies the condition, and so 11 * 5! * 4! out of 9! [email protected] bhai zara dekh lena if this matches whatever you got. Me off now...regardsscrabbler

@Logrhythm said:
a+p+b+q+c+r+d+s+e (a+b+c+d+e) - (p+q+r+s) = 0 or 11ktoo cumbersome to be done in office...

@viewpt said:
I too agree..mujhe toh abhi abhi kisi dusri company wale ka interview k liye call aaya mere boss k samne..@jain4444 bhai bauncer kara rahey ho ek k baad ek..

@insane.vodka said:
digits add up to 45 so case with 0 can be neglectedin 11k case difference cannot go to 22and to get difference 11 there is only one possibility of 17 28How to go beyond this...too weak at Probability

@Dexian said:
(9 5 2 1 ) ( 9 4 3 1 ) ( 8 6 2 1) ( 8 5 3 1) (8 4 3 2 ) (7 6 3 1) (7 5 3 2) (6 5 4 2)i m getting only 8 sets of 4 digit adding up to 17.... rest 5 will add up to 28..@scrabbler how r u getting 11 sets...plzz xplain.. which r d ones i m missing..

@scrabbler bhai

sum of all = 45

multiple of 11 les than 45

x = sum of even places
y = sum of odd places

x + y = 45
x - y = 11

y = 17 and x = 28

17 = (6 , 5 , 4 , 2) (1, 2, 5, 9), (1, 2, 6, 8 ), (1, 3, 4, 9), (1, 3, 5, 8 ), (1, 3, 6, 7), (1, 4, 5, 7), (2, 3, 4, 8 ), (2, 3, 5, 7), (1, 2, 3, 4, 7), (1, 2, 3, 5, 6)

11 ways

when sum of even places = 17
9*5!*4!

when sum of odd places = 17
2*5!*4!

total = 5!*4!*11

prob. = 11*5!*4!/9!