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# Anthony and Michael sit on the six-member board of directors

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Anthony and Michael sit on the six-member board of directors [#permalink]

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23 Jul 2007, 10:54
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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. 30%
b. 40%
c. 50%
d. 60%
e. 70%
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Joined: 04 Jun 2007
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23 Jul 2007, 10:59
lanter1 wrote:
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. 30%
b. 40%
c. 50%
d. 60%
e. 70%

B (40%) for me.

No. of sub-committees including Michael = 5C2 (M _ _) = 10
No. of sub-committees including Michael and Anthony = 4C1 (M A _) = 4
Answer = (4/10) * 100 = 40%
Senior Manager
Joined: 28 Jun 2007
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23 Jul 2007, 19:51
Can u explain how u arrived at 5C2 ??? Thanks in advance.
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Location: Singapore
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23 Jul 2007, 19:56
If the sub-committee already has michael on board, then the remaining two seats can be filled in 5C2 ways = 10

If the sub-committee has both michael and anthony, then the remaining seat can be filled 4 ways.

P = 4/10 = 2/5 = 40%
Senior Manager
Joined: 28 Jun 2007
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23 Jul 2007, 20:04
ywilfred wrote:
If the sub-committee already has michael on board, then the remaining two seats can be filled in 5C2 ways = 10

If the sub-committee has both michael and anthony, then the remaining seat can be filled 4 ways.

P = 4/10 = 2/5 = 40%

But Isnt the question a little different?

I would think they are asking for No. of Teams comprising both/ no. of Teams comprising one alone

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23 Jul 2007, 20:09
ioiio wrote:
ywilfred wrote:
If the sub-committee already has michael on board, then the remaining two seats can be filled in 5C2 ways = 10

If the sub-committee has both michael and anthony, then the remaining seat can be filled 4 ways.

P = 4/10 = 2/5 = 40%

But Isnt the question a little different?

I would think they are asking for No. of Teams comprising both/ no. of Teams comprising one alone

The question says 'what percent of all the possible subcommittees that include Michael also include Anthony? '

So it is looking for this percentage:

(All groupings that include Michael and Anthony)/(All groupings that include Michael) * 100%

So
All groupings that include Michael and Anthony -> you have to assume michael and anthony are already in the team, then the last seat can be just any one of the 4 remaining people

All groupings that include Michael -> you have to assume michale is already in the team, and the remaining two places can be filled by any 2 of the remaining 5 people, thus 5C2
Senior Manager
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Location: The 408
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23 Jul 2007, 20:17
sumande wrote:
lanter1 wrote:
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. 30%
b. 40%
c. 50%
d. 60%
e. 70%

B (40%) for me.

No. of sub-committees including Michael = 5C2 (M _ _) = 10
No. of sub-committees including Michael and Anthony = 4C1 (M A _) = 4
Answer = (4/10) * 100 = 40%

I'm new, so pardon the dumb n00b question. what do the letters C, M, and A represent?

Thanks!
Senior Manager
Joined: 28 Jun 2007
Posts: 456
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23 Jul 2007, 20:25
zakk wrote:
sumande wrote:
lanter1 wrote:
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. 30%
b. 40%
c. 50%
d. 60%
e. 70%

B (40%) for me.

No. of sub-committees including Michael = 5C2 (M _ _) = 10
No. of sub-committees including Michael and Anthony = 4C1 (M A _) = 4
Answer = (4/10) * 100 = 40%

I'm new, so pardon the dumb n00b question. what do the letters C, M, and A represent?

Thanks!

C is combination. Look up wiki for definition
M and A stand for Michael and Anthony.
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23 Jul 2007, 21:39
2/5

Assume anthony is selected in the committee, that leaves 5 people of which 2 other members are selected

total combinations of selecting 2 out of 5 = 5C2 = 10

Now, for michael to be one of the two there are 4C1 combinations = 4
i.e.
assume michael is fixed and the other can be anyone
4/10 = 2/5
23 Jul 2007, 21:39
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