official quant thread for cat08

[prince_chittu]

@implex
can u please elaborate your approach.

[STALWART]

Originally Posted by prince_chittu

There are infinite number of Red , Green , Yellow and Black colored balls . In how many ways one can choose 20 balls ?

22C2

Nah ..... It will be 23C3 ....

Sayonara

[MadMagga]

Can anyone please explain all the methods of solving different types of Remainder problems ?

[STALWART]

Originally Posted by selebratinglife

New problem!!
In how many ways can you place 6 different coloured balls in 3 identical baskets ?

Let A, B , C, D ,E and F are six differnt things.

Possible cases are ...

(0,0,6) -----> 1
(0,1,5) -----> 6C5 = 6
(0,2,4) -----> 6C4 = 15
(0,3,3) -----> 6C3/2! = 10
(1,1,4) -----> 6C4 = 15
(1,2,3) -----> 6C3 * 3C2 = 60
(2,2,2) -----> ( 6C2 * 4C2 )/3! = 15

Hence , adding 'em

Answer = 122


Sayonara

[prince_chittu]

@stalwart
(0,3,3) -----> 6C3/3! = 10

i think it shud be 6C3/2!.Can u please clarify?

[praveshnagra]

Originally Posted by STALWART

Nah ..... It will be 23C3 ....

Sayonara

Could u please post the solution along with it?

[hi.shivani]

Sum1 solve dis problem..

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers? (CAT 93)
a. 6666600
b. 6666660
c. 6666666
d. None of these

[poojadatta]

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers? (CAT 93)
a. 6666600
b. 6666660
c. 6666666
d. None of these


There is a formula in permutation combination for such type of prob

(n-1)!* (sum of terms)*(111...N times)

so here it is
4!*(1+3+5+7+9)*11111
24*25*11111=
6666600

Is answer rt?

[shivam_01]

Originally Posted by hi.shivani

Sum1 solve dis problem..

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers? (CAT 93)
a. 6666600
b. 6666660
c. 6666666
d. None of these

ok let c my solution
the number can be arranged in 120 ways
out of that 120 ways each number will be repeated 24 times so the unit digit will be (1+3+5+7+9)*24 which gives 0 as the digit and 60 as the carry
similarly the tens digit will be (1+3+5+7+9)*24 +60 which too gives 0 as the tens digit and the 66 as the remainder hence carrying on the same path we get a. 6666600.....

do u have the key let me know the solution >>>>>

[anil.iitkgp]

lemme join the thread...