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by vertigo05 » Wed Jul 29, 2009 10:29 am
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%
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by gauravgundal » Wed Jul 29, 2009 11:14 pm
Answer is 40 %

https://www.beatthegmat.com/very-tough-w ... 17504.html
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by vinayakdl » Thu Jul 30, 2009 5:07 am
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
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by raghavsarathy » Thu Jul 30, 2009 8:16 am
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%
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by vertigo05 » Thu Jul 30, 2009 10:27 am
i should have submitted this question in PS section. anyways, the OA is 40% can anybody explain it with proper justification?
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by kaulnikhil » Thu Jul 30, 2009 11:40 am
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


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by raghavsarathy » Thu Jul 30, 2009 4:43 pm
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
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by tohellandback » Thu Jul 30, 2009 9:41 pm
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%
The powers of two are bloody impolite!!
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