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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] New post 08 Dec 2006, 08:16
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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)20%
B)30%
C)40%
D)50%
E)60%
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 [#permalink] New post 08 Dec 2006, 10:41
possible subcommittees that include Michael = 6C2
possible subcommittees that include Michael also include Anthony= 6C1

(6C1/ 6C2) * 100 = 40%
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 [#permalink] New post 08 Dec 2006, 14:29
AK wrote:
possible subcommittees that include Michael = 6C2
possible subcommittees that include Michael also include Anthony= 6C1

(6C1/ 6C2) * 100 = 40%

Could you plz explain more. I am having difficulties following it. thanks
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Re: DS - board of directors [#permalink] New post 08 Dec 2006, 15:27
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?

(1st case)
M and A both are already member means we need to find commitees with 1 person of rest of 4

4C1 i.ie 4 ways to form a comitee.
2 comitees i.e 8 ways

(2nd case)
if michal is included already
5 people with 2 people comitee = 5C2=5*4/2=10
10 ways to form 1 comitte
togather 20 ways for 2 comittes

8/20= 2/5=40%
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 [#permalink] New post 08 Dec 2006, 18:14
arjsingh1976 wrote:
AK wrote:
possible subcommittees that include Michael = 6C2
possible subcommittees that include Michael also include Anthony= 6C1

(6C1/ 6C2) * 100 = 40%

Could you plz explain more. I am having difficulties following it. thanks


Damager has explained well.. I did a mistake while puting it here
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 [#permalink] New post 08 Dec 2006, 18:20
Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )
Answer = (4/10)*100 = 40%
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 [#permalink] New post 08 Dec 2006, 18:33
I am really confused by this ! It is not making sense
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 [#permalink] New post 08 Dec 2006, 20:02
I am really confused
I am usually good with combo problems


How do we get to these ?

Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Quote:
Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )
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 [#permalink] New post 08 Dec 2006, 21:21
how abt this...

one sub committee has 3 ppl nd u need to both anthony nd michael on it....

So..

3C2/6C2 or(+) 3C2/6C2...giving us 2/5 nd thereby 40%
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 [#permalink] New post 08 Dec 2006, 21:49
jamesrwright3 wrote:
I am really confused
I am usually good with combo problems


How do we get to these ?

Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Quote:
Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )


Comitee size =3
Michael has to be included. So he has already been chosen, i seat taken.
we need to find number of ways to choose members for remaing 2 seats.

Total number of people 6 - michel ( as he is already chosen = 5

Number of ways to choose 2 person out of 5 = 5C2= 5!/3!2!=5*4*3*2/( 3*2*2)===> 10

Similarly

When both michal and antony are chosen
total seats 3 . antony takes 1 snd micheal takes 1 . 1 seat remaining

Out of 6 people antony and michel alreasy chosen reming people 4

Number of ways to choose 1 peron out of 4 is 4C1=4!/3!=4
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 [#permalink] New post 09 Dec 2006, 00:06
Damager wrote:
jamesrwright3 wrote:
I am really confused
I am usually good with combo problems


How do we get to these ?

Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Quote:
Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )


Comitee size =3
Michael has to be included. So he has already been chosen, i seat taken.
we need to find number of ways to choose members for remaing 2 seats.

Total number of people 6 - michel ( as he is already chosen = 5

Number of ways to choose 2 person out of 5 = 5C2= 5!/3!2!=5*4*3*2/( 3*2*2)===> 10

Similarly

When both michal and antony are chosen
total seats 3 . antony takes 1 snd micheal takes 1 . 1 seat remaining

Out of 6 people antony and michel alreasy chosen reming people 4

Number of ways to choose 1 peron out of 4 is 4C1=4!/3!=4


Thanks. I got it now!
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 [#permalink] New post 09 Dec 2006, 08:04
arjsingh1976 wrote:
Damager wrote:
jamesrwright3 wrote:
I am really confused
I am usually good with combo problems


How do we get to these ?

Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Quote:
Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )


Comitee size =3
Michael has to be included. So he has already been chosen, i seat taken.
we need to find number of ways to choose members for remaing 2 seats.

Total number of people 6 - michel ( as he is already chosen = 5

Number of ways to choose 2 person out of 5 = 5C2= 5!/3!2!=5*4*3*2/( 3*2*2)===> 10

Similarly

When both michal and antony are chosen
total seats 3 . antony takes 1 snd micheal takes 1 . 1 seat remaining

Out of 6 people antony and michel alreasy chosen reming people 4

Number of ways to choose 1 peron out of 4 is 4C1=4!/3!=4


Thanks. I got it now!


Awesome explanation
Too bad we can't give people reputation on this board:)
Thanks!
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Re: DS - board of directors [#permalink] New post 10 Dec 2006, 16:57
arjsingh1976 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)20%
B)30%
C)40%
D)50%
E)60%


Is the answer 40% ? I am getting 20%. Well from what i understand it is

(No. Subcommittess including both Michael and Anthony) / (No. of all possible Subcommittees)

Which is 4C1 / 6C3 = 4/20 = 20%

What you guys are talking about is No. of subcommitees which has Michael, and not all possible subcommitees. So that number is 10

and that way u get 40%.

Please explain !
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Re: DS - board of directors [#permalink] New post 10 Dec 2006, 17:06
arjsingh1976 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?



Your equation should be:
subcomittees that include Michael+Anthony/no. with Michael + 2 others

Last edited by MBAlad on 10 Dec 2006, 17:25, edited 1 time in total.
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Re: DS - board of directors [#permalink] New post 10 Dec 2006, 17:15
das_go wrote:
arjsingh1976 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)20%
B)30%
C)40%
D)50%
E)60%


Is the answer 40% ? I am getting 20%. Well from what i understand it is

(No. Subcommittess including both Michael and Anthony) / (No. of all possible Subcommittees)

Which is 4C1 / 6C3 = 4/20 = 20%

What you guys are talking about is No. of subcommitees which has Michael, and not all possible subcommitees. So that number is 10

and that way u get 40%.

Please explain !



I solved this problem the same way as you did. However, I think the number of all possible subcommittees would be 10 instead of 20. (6C3)/2.
You are correct on the number of all possible sub committee if out of 6, you only have to pick 3 people. However, when the 6 people are SPLIT into 2 subgroupd, then when you pick out the 3 people, the other 3 left would also set one arrangement. so to avoid double counting, we need to divide that amount by 2. So, 20/2 = 10. the rest is the same.

Hope that makes sense.
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 [#permalink] New post 10 Dec 2006, 17:21
All possible with Michael = 5C2 = 5*4/2*1 = 10

As you've said, by selecting one individual subcommittee you leave a corresponding unique subcommittee so total number of unique subcommittees = 10

Note: We use 5C2 (not 6C3) because we are looking at total number of subcommittees with Michael, not the total number possible. Therefore, Michael has already taken the first seat so we look at the number of ways to choose two people from the remaining 5.
  [#permalink] 10 Dec 2006, 17:21
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