DS from manhattangmat

Answer is 40 %

https://www.beatthegmat.com/very-tough-w ... 17504.html

gauravgundal wrote:Answer is 40 %

https://www.beatthegmat.com/very-tough-w ... 17504.html

gaurav but no body ever posted an OA on that. I dont agree with 40%.

I think D:50%

vertigo05, it will be great if you can post OA here or on the original thread.


Vinayak

IMO - A

No of subcommitees in which both of them are together =4

Total number of subcommmitees = 6C3 = 20

Hence required probablity = 4/20 = 20%

i should have submitted this question in PS section. anyways, the OA is 40% can anybody explain it with proper justification?

let us name the members
Mark Anthony X1 X2 X3 X4

now when mark and anthony are together

MARK ANthony X1 ---- X2 X3 X4
simlarly replace X1 by X2
MARK ANthony X2 ---- X1 X3 X4
now replace X2 by X3
MARK ANthony X3 ---- X2 X1 X4
now replace X3 by X4
MARK ANthony X4 ---- X2 X3 X1

these are the cases when both are together --4 such cases

now when they not together
M X1 X2 ---------------A X3 X4
M X1 X3---------------A X2 X4
M X1 x4 --------------A X2 X3

M X2 X3 ------------- A X1 X4
M X2 X4 ----------------A X1 X3

M X3 X4 ----------------A X1 X2
these comes to 6

so favourable cases = 4/10*100= 40

raghavsarathy wrote:IMO - A

No of subcommitees in which both of them are together =4

Total number of subcommmitees = 6C3 = 20

Hence required probablity = 4/20 = 20%

Why can't we say that the total number of subcommitees =6C3 ?

I am not able to understand clearly.

Kindly help

raghavsarathy wrote:

raghavsarathy wrote:IMO - A

No of subcommitees in which both of them are together =4

Total number of subcommmitees = 6C3 = 20

Hence required probablity = 4/20 = 20%

Why can't we say that the total number of subcommitees =6C3 ?

I am not able to understand clearly.

Kindly help

6C3 is wrong because the two groups are similar in the sense that when you select 3 out of 6, the other groups member are automatically getting selected.
it's like this, how many ways you can separate AB into 2 different alphabets.
its not two ways. Only one way. alphabets A and B are different but the groups you are dividing them into are similar
back to problem.
total number of ways =6C3/2!= 10
total numbers ways when anthony and michael are together =4C1=4 ways
4/10= 40%